Solving Quadratic Equations by Completing the Square: Combination with analytic geometry

Examples with solutions for Solving Quadratic Equations by Completing the Square: Combination with analytic geometry

Exercise #1

 

Below is an equation for a circle:

x28x+y26y=24 x^2-8x+y^2-6y=24

Solve the equation by completing the square in order to determine

the centre of the circle as well as the radius.

Step-by-Step Solution

Let's recall first that the equation of a circle with center at pointO(xo,yo) O(x_o,y_o) and radius R R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equation and examine it:

x28x+y26y=24 x^2-8x+y^2-6y=24

We'll try to give this equation a form identical to the circle equation, meaning - we'll ensure that the right side contains the sum of two squared binomial expressions, one for x and one for y, we'll do this using the "completing the square" method:

For this, first let's recall the binomial square formulas:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2

and let's deal separately with the part of the equation related to x (underlined):

x28x+y26y=24 \underline{ x^2-8x}+y^2-6y=24

Let's continue, for convenience and clarity of discussion - let's separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the first two terms in the binomial square formula (we'll choose the subtraction form of the binomial square formula since the first-degree term in the expression we're dealing with8x 8x has a negative sign):

x28xc22cd+d2x22x4c22cd+d2 \underline{ x^2-8x} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\ We can notice that compared to the binomial square formula (from the right side of the blue press in the previous calculation) we are actually making the analogy:

{xc4d \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4\textcolor{blue}{\leftrightarrow}d \end{cases}

Therefore, we can identify that if we want to get from these two terms (underlined in the calculation) a binomial square form,

We'll need to add to these two terms the term42 4^2 , but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

Meaning, we'll add and subtract the term (or expression) we need to "complete" to a binomial square form,

In the following calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next, we'll put into binomial square form the appropriate expression (demonstrated with colors) and in the final stage we'll further simplify the expression:

x22x4x22x4+4242x22x4+4216(x4)216 x^2-2\cdot x\cdot 4\\ x^2-2\cdot x\cdot 4\underline{\underline{+4^2-4^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}+\textcolor{green}{4}^2-16\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4})^2-16}\\ Let's summarize the development stages so far for the x-related expression, we'll do this now within the given equation:

x28x+y26y=24x22x4+y26y=24x22x4+4242+y26y=24(x4)216+y26y=24 x^2-8x+y^2-6y=24 \\ x^2-2\cdot x\cdot 4+y^2-6y=24 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4}\underline{\underline{+\textcolor{green}{4}^2-4^2}}+y^2-6y=24\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4})^2-16+y^2-6y=24\\ We'll continue and perform an identical process for the y-related terms in the resulting equation:

(x4)216+y26y=24(x4)216+y22y3=24(x4)216+y22y3+3232=24(x4)216+y22y3+329=24(x4)216+(y3)29=24(x4)2+(y3)2=49 (x-4)^2-16+\underline{y^2-6y}=24\\ \downarrow\\ (x-4)^2-16+\underline{y^2-2\cdot y\cdot 3}=24\\ (x-4)^2-16+\underline{y^2-2\cdot y\cdot 3\underline{\underline{+3^2-3^2}}}=24\\ \downarrow\\ (x-4)^2-16+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9}=24\\ \downarrow\\ (x-4)^2-16+(\textcolor{red}{y}-\textcolor{green}{3})^2-9=24\\ \boxed{(x-4)^2+(y-3)^2=49}

In the last stage, we moved the free numbers to the other side and grouped similar terms,

Now that we've transformed the given circle equation into the form of the general circle equation mentioned earlier, we can easily extract both the center of the given circle and its radius from the given equation:

(xxo)2+(yyo)2=R2(x4)2+(y3)2=49 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4})^2+(y-\textcolor{orange}{3})^2=\underline{\underline{49}}

Therefore we can conclude that the circle's center is at point:(xo,yo)(4,3) \boxed{(x_o,y_o)\leftrightarrow(4,3)} and extract the circle's radius by solving a simple equation:

R2=49/R=7 R^2=49\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=7}

(as we remember that the circle's radius by definition is a distance from any point on the circle to the circle's center - is positive),

Therefore, the correct answer is answer C.

Answer

(4,3),R=7 (4,3),\hspace{6pt} R=7

Exercise #2

Below is the equation for a circle:

27x2+4x+27y2+87y=227 \frac{2}{7}x^2+4x+ \frac{2}{7}y^2+ \frac{8}{7}y= \frac{22}{7}

Solve the equation by completing the square in order to find the centre point of the circle as well as the radius

Step-by-Step Solution

Let's recall first that the equation of a circle with center at point

O(xo,yo) O(x_o,y_o)

and radius R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equation and examine it:

27x2+4x+27y2+87y=227 \frac{2}{7}x^2+4x+ \frac{2}{7}y^2+ \frac{8}{7}y= \frac{22}{7}

First, let's note that the form of the given equation is not exactly identical to the general circle equation form mentioned, which we know, because the coefficients of the squared terms are not 1, however they are equal to each other, and therefore this equation must represent a circle, but in order to get its characteristics we must first divide both sides of the equation by the coefficient of the squared terms,

27 \frac{2}{7}

, however since we're dealing with a fraction - we prefer to multiply instead of divide, meaning - we'll multiply both sides of the equation by

72 \frac{7}{2}

Next we'll reduce the fraction multiplications in the expression:

27x2+4x+27y2+87y=227/727227x2+724x+7227y2+7287y=72227x2+14x+y2+4y=11 \frac{2}{7}x^2+4x+ \frac{2}{7}y^2+ \frac{8}{7}y= \frac{22}{7} \hspace{6pt}\text{/}\cdot \frac{7}{2} \\ \downarrow\\ \frac{7}{2}\cdot \frac{2}{7}x^2+\frac{7}{2}\cdot4x+ \frac{7}{2}\cdot\frac{2}{7}y^2+ \frac{7}{2}\cdot\frac{8}{7}y= \frac{7}{2}\cdot\frac{22}{7} \\ \downarrow\\ x^2+14x+y^2+4y=11

We have thus obtained an equation equivalent to the given one, in the correct form - meaning where the coefficients of the squared terms are 1, from here we can continue and get the circle's characteristics:

We'll try to give this equation a form identical to the circle equation, meaning - we'll ensure that on its right side there will be the sum of two squared binomial expressions, one for x and one for y, we'll do this using the "completing the square" method:

For this, first let's recall again the perfect square trinomial formulas:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2

And let's deal separately with the part of the equation related to x in the equation (underlined):

x2+14x+y2+4y=11x2+14x+y2+4y=11 x^2+14x+y^2+4y=11 \\ \underline{ x^2+14x}+y^2+4y=11

Let's continue, for convenience and clarity of discussion - let's separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the first two terms in the perfect square formula (we'll choose the addition form of the perfect square trinomial since the first-degree term in the expression we're dealing with 14x 14x has a positive sign):

x2+14xc2+2cd+d2x2+2x7c2+2cd+d2 \underline{ x^2+14x} \textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{7}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\

One can notice that compared to the perfect square formula (from the blue box on the right in the previous calculation) we are actually making the analogy:

{xc7d \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 7\textcolor{blue}{\leftrightarrow}d \end{cases}

Therefore, we identify that if we want to get from these two terms (underlined in the calculation) a perfect square trinomial form,

we'll need to add to these two terms the term72 7^2 , but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

meaning - we'll add and subtract the term (or expression) we need to "complete" to a perfect square trinomial,

In the next calculation the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

x2+2x7x2+2x7+7272x2+2x7+7249(x+7)249 x^2+2\cdot x\cdot 7\\ x^2+2\cdot x\cdot 7\underline{\underline{+7^2-7^2}}\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot \textcolor{green}{7}+\textcolor{green}{7}^2-49\\ \downarrow\\ \boxed{ (\textcolor{red}{x}+\textcolor{green}{7})^2-49}\\

Let's summarize the development stages so far for the expression related to x, we'll do this now within the given equation:

x2+14x+y2+4y=11x2+2x7+y2+4y=11x2+2x7+7272+y2+4y=11(x+7)249+y2+4y=11 x^2+14x+y^2+4y=11 \\ x^2+2\cdot x\cdot 7+y^2+4y=11 \\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{7}\underline{\underline{+\textcolor{green}{7}^2-7^2}}+y^2+4y=11\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{7})^2-49+y^2+4y=11\\

We'll continue and perform an identical process also for the terms related to y in the equation we obtained:

(x+7)249+y2+4y=11(x+7)249+y2+2y2=11(x+7)249+y2+2y2+2222=11(x+7)249+y2+2y2+224=11(x+7)249+(y+2)24=11(x+7)2+(y+2)2=64 (x+7)^2-49+\underline{y^2+4y}=11\\ \downarrow\\ (x+7)^2-49+\underline{y^2+2\cdot y\cdot 2}=11\\ (x+7)^2-49+\underline{y^2+2\cdot y\cdot 2\underline{\underline{+2^2-2^2}}}=11\\ \downarrow\\ (x+7)^2-49+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=11\\ \downarrow\\ (x+7)^2-49+(\textcolor{red}{y}+\textcolor{green}{2})^2-4=11\\ \boxed{ (x+7)^2+(y+2)^2=64}

In the last stage we moved the free numbers to the other side and grouped similar terms,

Now that we've changed the given circle equation to the form of the general circle equation mentioned earlier, we can simply extract both the given circle's center and its radius from the given equation:

(xxo)2+(yyo)2=R2(x+7)2+(y+2)2=64(x(7))2+(y(2))2=64 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x+\textcolor{purple}{7})^2+(y+\textcolor{orange}{2})^2=\underline{\underline{64}}\\ \downarrow\\ \big(x-(\textcolor{purple}{-7})\big)^2+\big(y-(\textcolor{orange}{-2})\big)^2=\underline{\underline{64}}\\

Therefore we can conclude that the circle's center point is:

(xo,yo)(7,2) \boxed{(x_o,y_o)\leftrightarrow(-7,-2)}

And extract the circle's radius by solving a simple equation:

R2=64/R=8 R^2=64\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=8}

(as we remember that the circle's radius by definition is a distance from any point on the circle to the circle's center - is positive),

Therefore, the correct answer is answer C.

Answer

(7,2),R=8 (-7,-2),\hspace{6pt} R=8

Exercise #3

Below is an equation for a circle:

3x2+30x+3y290y=333 3x^2+30x+3y^2-90y=333

Solve the equation by completing the square in order to determine

the centre of the circle as well as the radius:

Step-by-Step Solution

Let's recall first that the equation of a circle with center at point O(xo,yo) O(x_o,y_o) and radius R R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equation and examine it:

3x2+30x+3y290y=333 3x^2+30x+3y^2-90y=333

Let's first notice that the form of the given equation is not completely identical to the general circle equation form we mentioned, which we know, because the coefficients of the squared terms are not 1, however they are equal to each other, and therefore this equation must represent a circle, but in order to get its characteristics we must first divide both sides of the equation by the coefficient of the squared terms, 3:

3x2+30x+3y290y=333/:3x2+10x+y230y=111 3x^2+30x+3y^2-90y=333\hspace{6pt}\text{/}:3\\ \downarrow\\ x^2+10x+y^2-30y=111

We have thus obtained an equation equivalent to the given equation, in the correct form - meaning - where the coefficients of the squared terms are 1, from here we can continue and get the circle's characteristics:

Let's try to give this equation a form identical to the circle equation form, meaning - we'll ensure that the right side has the sum of two squared binomial expressions, one for x and one for y, we'll do this using the "completing the square" method:

For this, first let's recall again the binomial square formulas:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2

and let's deal separately with the part of the equation related to x (underlined):

x2+10x+y230y=111 \underline{x^2+10x}+y^2-30y=111

Let's continue, for convenience and clarity of discussion - let's separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the first two terms in the binomial square formula (we'll choose the addition form of the binomial square formula since the first-degree term in the expression we're dealing with 10x 10x has a positive sign):

x2+10xc2+2cd+d2x2+2x5c2+2cd+d2 \underline{ x^2+10x} \textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{5}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\ One can notice that compared to the binomial square formula (on the right side of the blue press in the previous calculation) we are actually making the analogy:

{xc5d \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 5\textcolor{blue}{\leftrightarrow}d \end{cases}

Therefore, we can identify that if we want to get from these two terms (underlined in the calculation) a binomial square form,

We'll need to add to these two terms the term52 5^2 , but we don't want to change the value of the expression in question, so we'll also subtract this term from the expression,

Meaning - we'll add and subtract the term (or expression) we need to "complete" to a binomial square form,

In the next calculation the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next - we'll put into binomial square form the appropriate expression (demonstrated using colors) and in the final stage we'll further simplify the expression:

x2+2x5x2+2x5+5252x2+2x5+5225(x+5)225 x^2+2\cdot x\cdot 5\\ x^2+2\cdot x\cdot 5\underline{\underline{+5^2-5^2}}\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot \textcolor{green}{5}+\textcolor{green}{5}^2-25\\ \downarrow\\ \boxed{ (\textcolor{red}{x}+\textcolor{green}{5})^2-25}\\ Let's summarize the development stages so far for the expression related to x, we'll do this now within the given equation:

x2+10x+y230y=111x2+2x5+y230y=111x2+2x5+5252+y230y=111(x+5)225+y230y=111 x^2+10x+y^2-30y=111 \\ x^2+2\cdot x\cdot 5+y^2-30y=111 \\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{5}\underline{\underline{+\textcolor{green}{5}^2-5^2}}+y^2-30y=111\\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{5})^2-25+y^2-30y=111\\ We'll continue and perform an identical process also for the expressions related to y in the equation we got:

(x+5)225+y230y=111(x+5)225+y22y15=111(x+5)225+y22y15+152152=111(x+5)225+y22y15+152225=111(x+5)225+(y15)2225=111(x+5)2+(y15)2=361 (x+5)^2-25+\underline{y^2-30y}=111\\ \downarrow\\ (x+5)^2-25+\underline{y^2-2\cdot y\cdot 15}=111\\ (x+5)^2-25+\underline{y^2-2\cdot y\cdot 15\underline{\underline{+15^2-15^2}}}=111\\ \downarrow\\ (x+5)^2-25+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{15}+\textcolor{green}{15}^2-225}=111\\ \downarrow\\ (x+5)^2-25+(\textcolor{red}{y}-\textcolor{green}{15})^2-225=111\\ \boxed{ (x+5)^2+(y-15)^2=361}

In the last stage we moved the free numbers to the other side and entered similar terms,

Now that we've changed the given circle equation to the form of the general circle equation mentioned earlier, we can simply extract both the center of the given circle and its radius:

(xxo)2+(yyo)2=R2(x+5)2+(y15)2=361(x(5))2+(y15)2=361 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x+\textcolor{purple}{5})^2+(y-\textcolor{orange}{15})^2=\underline{\underline{361}}\\ \downarrow\\ (x-(\textcolor{purple}{-5}))^2+(y-\textcolor{orange}{15})^2=\underline{\underline{361}}\\ Therefore we can conclude that the circle's center is at point:(xo,yo)(5,15) \boxed{(x_o,y_o)\leftrightarrow(-5,15)}

and extract the circle's radius by solving a simple equation:

R2=361/R=19 R^2=361\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=19} (where we remember that the circle's radius by definition is a distance from any point on the circle to the circle's center - is positive),

Therefore, the correct answer is answer D.

Answer

(5,15),R=19 (-5,15),\hspace{6pt} R=19

Exercise #4

A circle has the equation:

x26x+y24y=7 x^2-6x+y^2-4y=7

Use the completing the square method to find the center of the circle and its radius.

Step-by-Step Solution

Let's first recall that the equation of a circle with center at point O(xo,yo) O(x_o,y_o) and radius R R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equation and examine it:

x26x+y24y=7 x^2-6x+y^2-4y=7

We'll try to give this equation a form identical to the circle equation, meaning - we'll ensure that the right side contains the sum of two squared binomial expressions, one for x and one for y. We'll do this using the "completing the square" method:

For this, first let's recall the formulas for squared binomial expressions:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2

and let's deal separately with the x-related part of the equation (underlined):

x26x+y24y=7 \underline{ x^2-6x}+y^2-4y=7

Let's continue, for convenience and clarity - let's separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the first two terms in the squared binomial formula (we'll choose the subtraction form of the squared binomial formula since the first-degree term in the expression we're dealing with 6x 6x has a negative sign):

x26xc22cd+d2x22x3c22cd+d2 \underline{ x^2-6x} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\ We can notice that compared to the squared binomial formula (from the blue box on the right in the previous calculation) we're actually making the analogy:

{xc3d \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 3\textcolor{blue}{\leftrightarrow}d \end{cases}

Therefore, we can identify that if we want to get a squared binomial form from these two terms (underlined in the calculation),

We'll need to add to these two terms the term32 3^2 , but we don't want to change the value of the expression, so we'll also subtract this term from the expression,

In other words, we'll add and subtract the term (or expression) needed to "complete" the squared binomial form,

In the next calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Next, we'll put into squared binomial form the appropriate expression (demonstrated with colors) and in the final step we'll simplify the expression further:

x22x3x22x3+3232x22x3+329(x3)29 x^2-2\cdot x\cdot 3\\ x^2-2\cdot x\cdot 3\underline{\underline{+3^2-3^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{3})^2-9}\\ Let's summarize the development steps so far for the x-related expression, we'll do this now within the given equation:

x26x+y24y=7x22x3+y24y=7(x)22x3+3232+y24y=7(x3)29+y24y=7 x^2-6x+y^2-4y=7 \\ x^2-2\cdot x\cdot 3+y^2-4y=7 \\ (\textcolor{red}{x})^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{3}\underline{\underline{+\textcolor{green}{3}^2-3^2}}+y^2-4y=7\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{3})^2-9+y^2-4y=7\\ We'll continue and perform an identical process for the y-related terms in the resulting equation:

(x3)29+y24y=7(x3)29+y22y2=7(x3)29+y22y2+2222=7(x3)29+y22y2+224=7(x3)29+(y2)24=7(x3)2+(y2)2=20 (x-3)^2-9+\underline{y^2-4y}=7\\ \downarrow\\ (x-3)^2-9+\underline{y^2-2\cdot y\cdot 2}=7\\ (x-3)^2-9+\underline{y^2-2\cdot y\cdot 2\underline{\underline{+2^2-2^2}}}=7\\ \downarrow\\ (x-3)^2-9+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=7\\ \downarrow\\ (x-3)^2-9+(\textcolor{red}{y}-\textcolor{green}{2})^2-4=7\\ \boxed{(x-3)^2+(y-2)^2=20}

In the last step, we moved the free numbers to the other side and grouped similar terms,

Now that we've transformed the given circle equation into the general circle equation form mentioned earlier, we can easily extract both the center of the given circle and its radius from the given equation:

(xxo)2+(yyo)2=R2(x3)2+(y2)2=20 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{3})^2+(y-\textcolor{orange}{2})^2=\underline{\underline{20}}

Therefore we can conclude that the circle's center is at point:(xo,yo)(3,2) \boxed{(x_o,y_o)\leftrightarrow(3,2)} and extract the circle's radius by solving a simple equation:

R2=20/R=20 R^2=20\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=\sqrt{20}}

(remembering that the circle's radius by definition is a distance from any point on the circle to its center - is positive),

Therefore, the correct answer is answer D.

Answer

(3,2),R=20 (3,2),\hspace{6pt} R=\sqrt{20}

Exercise #5

In the following diagram, mark points O O and M M in the plane:

x222x+y2+6y=114Ox28ax+y2+4ay=11a2M x^2-22x+y^2+6y=-114\leftrightarrow O\\ x^2-8ax+y^2+4ay=-11a^2\leftrightarrow M\\

Given the coordinates of the points OM OM relative to x

Calculate a a , and the radius of the circle centered on it:

RM R_M .

Step-by-Step Solution

Let's recall first that the equation of a circle with center at point:

O(xo,yo) O(x_o,y_o)

and radiusR R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equations and examine it:

x222x+y2+6y=114Ox28ax+y2+4ay=11a2M x^2-22x+y^2+6y=-114\leftrightarrow O\\ x^2-8ax+y^2+4ay=-11a^2\leftrightarrow M\\ We have two circle equations, from which we can extract the centers of the circles and their radii,

We will do this using "completing the square", let's start with the first circle equation (with center at point O O ):

x222x+y2+6y=114x22x11+y2+2y3=114(x11)2112+(y+3)232=114(x11)2+(y+3)2=16O(11,3),RO=4 x^2-22x+y^2+6y=-114\\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{11}+y^2+2\cdot y\cdot \textcolor{red}{3}=-114 \\ \downarrow\\ (x-\textcolor{blue}{11})^2-\textcolor{blue}{11}^2+(y+\textcolor{red}{3})^2-\textcolor{red}{3}^2=-114\\ (x-11)^2+(y+3)^2=16\\ \rightarrow\boxed{O(11,-3),\hspace{4pt}R_O=4}

We'll continue with the same process for the second circle (with center at point M M ):

x28ax+y2+4ay=11a2x22x4a+y2+2y2a=11a2(x4a)2(4a)2+(y+2a)2(2a)2=11a2(x4a)216a2+(y+2a)24a2=11a2(x4a)2+(y+2a)2=9a2(x4a)2+(y+2a)2=(3a)2M(4a,2a),RM=(+)3a x^2-8ax+y^2+4ay=-11a^2\\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{4a}+y^2+2\cdot y\cdot \textcolor{red}{2a}=-11a^2\\ \downarrow\\ (x-\textcolor{blue}{4a})^2-(\textcolor{blue}{4a})^2+(y+\textcolor{red}{2a})^2-(\textcolor{red}{2a})^2=-11a^2\\ (x-4a)^2-16a^2+(y+2a)^2-4a^2=-11a^2\\ (x-4a)^2+(y+2a)^2=9a^2\\ (x-4a)^2+(y+2a)^2=(3a)^2\\ \rightarrow\boxed{M(4a,-2a),\hspace{4pt}R_M=(\textcolor{red}{+})3a}

(Note that since it's given that: a>0 , we chose the positive sign in the radius expression we found earlier)

Now that we have expressions for the centers of the circles, we'll use the fact that the line segment between the centersOM OM (the line connecting the centers of the two circles) is parallel to the x-axis,

Remember that for any point along a line parallel to the x-axis, the y-coordinate remains constant.

Therefore, it must be true that:

yO=yM y_O=y_M , therefore, we'll equate the y-coordinates of the circle centers we found and solve this equation for a a :

{O(11,3)M(4a,2a)yO=yM2a=3a=32 \begin{cases} O(11,-3)\\ M(4a,-2a) \end{cases}\leftrightarrow y_O=y_M\\ \downarrow\\ -2a=-3\\ \downarrow\\ \boxed{a=\frac{3} {2}}

We'll continue and calculate using the parameter value we found and using the radius expression M M , that we found earlier, the value of this circle's radius:

RM=3aa=32RM=332=92 R_M=3a\leftrightarrow a=\frac{3}{2}\\ \downarrow\\ \boxed{R_M=3\cdot\frac{3}{2}=\frac{9}{2}}

Therefore, the correct answer is answer C.

Answer

a=32,RM=92 a=\frac{3}{2},\hspace{4pt}R_M=\frac{9}{2}

Exercise #6

A circle has the following equation:
x28ax+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2

Point O is its center and is in the second quadrant (a0 a\neq0 )


Use the completing the square method to find the center of the circle and its radius in terms of a a .

Step-by-Step Solution

 Let's recall that the equation of a circle with its center at O(xo,yo) O(x_o,y_o) and its radius R R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2 Now, let's now have a look at the equation for the given circle:

x28ax+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2
We will try rearrange this equation to match the circle equation, or in other words we will ensure that on the left side is the sum of two squared binomial expressions, one for x and one for y.

We will do this using the "completing the square" method:

Let's recall the short formula for squaring a binomial:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2 We'll deal separately with the part of the equation related to x in the equation (underlined):

x28ax+y2+10ay=5a2 \underline{ x^2-8ax}+y^2+10ay=-5a^2

We'll isolate these two terms from the equation and deal with them separately.

We'll present these terms in a form similar to the form of the first two terms in the shortcut formula (we'll choose the subtraction form of the binomial squared formula since the term in the first power we are dealing with is8ax 8ax , which has a negative sign):

x28axc22cd+d2x22x4ac22cd+d2 \underline{ x^2-8ax} \textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\ Notice that compared to the short formula (which is on the right side of the blue arrow in the previous calculation), we are actually making the comparison:

{xc4ad \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4a\textcolor{blue}{\leftrightarrow}d \end{cases} Therefore, if we want to get a squared binomial form from these two terms (underlined in the calculation), we will need to add the term(4</span><spanclass="katex">a)2 (4</span><span class="katex">a)^2 , but we don't want to change the value of the expression, and therefore we will also subtract this term from the expression.

That is, we will add and subtract the term (or expression) we need to "complete" to the binomial squared form,

In the following calculation, the "trick" is highlighted (two lines under the term we added and subtracted from the expression),

Next, we'll put the expression in the squared binomial form the appropriate expression (highlighted with colors) and in the last stage we'll simplify the expression:

x22x4ax22x4a+(4a)2(4a)2x22x4a+(4a)216a2(x4a)216a2 x^2-2\cdot x\cdot 4a\\ x^2-2\cdot x\cdot4a\underline{\underline{+(4a)^2-(4a)^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4a}+(\textcolor{green}{4a})^2-16a^2\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2}\\ Let's summarize the steps we've taken so far for the expression with x.

We'll do this within the given equation:

x28ax+y2+10ay=5a2x22x4a+(4a)2(4a)2+y2+10ay=5a2(x4a)216a2+y2+10ay=5a2 x^2-8ax+y^2+10ay=-5a^2 \\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4a}\underline{\underline{+\textcolor{green}{(4a)}^2-(4a)^2}}+y^2+10ay=-5a^2\\ \downarrow\\ (\textcolor{red}{x}-\textcolor{green}{4a})^2-16a^2+y^2+10ay=-5a^2\\ We'll continue and do the same thing for the expressions with y in the resulting equation:

(Now we'll choose the addition form of the squared binomial formula since the term in the first power we are dealing with 10ay 10ay has a positive sign)

(x4a)216a2+y2+10ay=5a2(x4a)216a2+y2+2y5a=5a2(x4a)216a2+y2+2y5a+(5a)2(5a)2=5a2(x4a)216a2+y2+2y5a+(5a)225a2=5a2(x4a)216a2+(y+5a)225a2=5a2(x4a)2+(y+5a)2=36a2 (x-4a)^2-16a^2+\underline{y^2+10ay}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a}=-5a^2\\ (x-4a)^2-16a^2+\underline{y^2+2\cdot y \cdot 5a\underline{\underline{+(5a)^2-(5a)^2}}}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{5a}+\textcolor{green}{(5a)}^2-25a^2}=-5a^2\\ \downarrow\\ (x-4a)^2-16a^2+(\textcolor{red}{y}+\textcolor{green}{5a})^2-25a^2=-5a^2\\ \boxed{(x-4a)^2+(y+5a)^2=36a^2} In the last step, we move the free numbers to the second side and combine like terms.

Now that the given circle equation is in the form of the general circle equation mentioned earlier, we can easily extract both the center of the given circle and its radius:

(xxo)2+(yyo)2=R2(x4a)2+(y+5a)2=36a2(x4a)2+(y(5a))2=36a2 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ (x-\textcolor{purple}{4a})^2+(y+\textcolor{orange}{5a})^2=\underline{\underline{36a^2}}\\ \downarrow\\ (x-\textcolor{purple}{4a})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{5a}))^2=\underline{\underline{36a^2}}\\

In the last step, we made sure to get the exact form of the general circle equation—that is, where only subtraction is performed within the squared expressions (emphasized with an arrow)

Therefore, we can conclude that the center of the circle is at:O(xo,yo)O(4a,5a) \boxed{O(x_o,y_o)\leftrightarrow O(4a,-5a)} and extract the radius of the circle by solving a simple equation:

R2=36a2/R=±6a R^2=36a^2\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=\pm6a}

Remember that the radius of the circle, by its definition is the distance between any point on the diameter and the center of the circle. Since it is positive, we must disqualify one of the options we got for the radius.

To do this, we will use the remaining information we haven't used yet—which is that the center of the given circle O is in the second quadrant.

That is:

O(x_o,y_o)\leftrightarrow x_o<0,\hspace{4pt}y_o>0 (Or in words: the x-value of the circle's center is negative and the y-value of the circle's center is positive)

Therefore, it must be true that:

\begin{cases} x_o<0\rightarrow (x_o=4a)\rightarrow 4a<0\rightarrow\boxed{a<0}\\ y_o>0\rightarrow (y_o=-5a)\rightarrow -5a>0\rightarrow\boxed{a<0} \end{cases}

We concluded that a<0 and since the radius of the circle is positive we conclude that necessarily:

R=6a \rightarrow \boxed{R=-6a} Let's summarize:

O(4a,5a),R=6a \boxed{O(4a,-5a), \hspace{4pt}R=-6a} Therefore, the correct answer is answer d. 

Answer

O(4a,5a),R=6a O(4a,-5a),\hspace{4pt}R=-6a

Exercise #7

x210ax+y24yb=20ab x^2-10ax+y^2-4yb=-20ab

The above equation represents a circle

If the parameters are

a,b a,\hspace{2pt}b

then the solution depends on the following:

y=x y=x

Step-by-Step Solution

Let's recall first that the equation of a circle with center at point:

O(xo,yo) O(x_o,y_o)

and its radiusR R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equation and examine it:

x210ax+y24yb=20ab x^2-10ax+y^2-4yb=-20ab

We can extract from it the circle's center and its (square) radius,

We'll do this using "completing the square":

x210ax+y24yb=20abx22x5a+y22y2b=20ab(x5a)2(5a)2+(y2b)2(2b)2=20ab(x5a)2+(y2b)2=(5a)220ab+(2b)2O(5a,2b),RO2=(5a)220ab+(2b)2 x^2-10ax+y^2-4yb=-20ab \\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{5a}+y^2-2\cdot y\cdot \textcolor{red}{2b}=-20ab \\ \downarrow\\ (x-\textcolor{blue}{5a})^2-(\textcolor{blue}{5a})^2+(y-\textcolor{red}{2b})^2-(\textcolor{red}{2b})^2=-20ab\\ (x-5a)^2+(y-2b)^2=(5a)^2-20ab+(2b)^2\\ \rightarrow\boxed{O(5a,2b),\hspace{4pt}R_O^2=(5a)^2-20ab+(2b)^2}

Note that for the circle's radius squared we got an algebraic expression that includes both parameters from the problem,

Note additionally that this expression can be factored into a squared binomial (using the perfect square formula):

RO2=(5a)220ab+(2b)2RO2=(5a)225a2b+(2b)2RO2=(5a2b)2 R_O^2=(5a)^2-20ab+(2b)^2 \\ \downarrow\\ R_O^2=(5a)^2-2\cdot5a\cdot2b+(2b)^2\\ \boxed{ R_O^2=(5a-2b)^2}

To summarize, we found that the center of the given circle represented by the equation and its radius (squared) are:

O(5a,2b),RO2=(5a2b)2 \rightarrow\boxed{O(5a,2b),\hspace{4pt}R_O^2=(5a-2b)^2}

Now let's note that there is a constraint on the parameters in the problem, let's look at the radius squared expression: RO2 R_O^2 ,

and note that although this expression: (5a2b)2 (5a-2b)^2 must be non-negative (since it's an expression that is squared) meaning there's no contradiction with the fact that it equals the radius squared,

still- it could be zero, which would contradict the initial given- stating that the given equation represents a circle equation, (and therefore it's impossible that the radius derived from it would be zero),

meaning we must require that:

RO05a2b05a2b R_O\neq0\\ \downarrow\\ 5a-2b\neq0\\ \downarrow\\ \boxed{5a\neq2b}

Let's return now to the expression for the circle's center that we got and note that we can conclude from the constraint we found on the parameters that:

O(xo,yo)O(5a,2b)5a2bxoyo O(x_o,y_o)\leftrightarrow O(5a,2b)\leftrightarrow 5a\neq2b\\ \downarrow\\ \boxed{x_o\neq y_o}

meaning- it's impossible for the circle's center point to have identical x and y coordinates!

However- if we assume that the circle's center point indeed lies on the line: y=x y=x then it must satisfy it, meaning it must be true that:

yo=xo \textcolor{red}{\boxed{y_o=x_o}}

But of course as mentioned before this is a contradiction to the constraint we found earlier(!!)

(which came from requiring that the circle's radius cannot be zero)

Therefore we conclude that there are no parametersa,b a,\hspace{2pt}b for which the circle's center (represented by the given equation) lies on the line: y=x y=x .

Therefore- the correct answer is answer C.

Answer

a,b a,\hspace{2pt}b do not exist

Exercise #8

Below is the equation for a circle:

x2+2ax+y22by=0 x^2+2ax+y^2-2by=0

is the centre point O O ,

We will draw the y-axis, given the function:

a<0

Which of the following statements is correct?

Step-by-Step Solution

Let's recall first that the equation of a circle with center at point:

O(xo,yo) O(x_o,y_o)

and radiusR R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Let's now return to the problem and the given circle equation and examine it:

x2+2ax+y22by=0 x^2+2ax+y^2-2by=0

we can extract from it the center of the circle and its (squared) radius,

we'll do this using "completing the square":

x2+2ax+y22yb=0x2+2xa+y22yb=0(xa)2a2+(yb)2b2=0(x+a)2+(yb)2=a2+b2O(a,b),RO2=a2+b2 x^2+2ax+y^2-2yb=0 \\ \downarrow\\ x^2+2\cdot x\cdot \textcolor{blue}{a}+y^2-2\cdot y\cdot \textcolor{red}{b}=0 \\ \downarrow\\ (x-\textcolor{blue}{a})^2-\textcolor{blue}{a}^2+(y-\textcolor{red}{b})^2-\textcolor{red}{b}^2=0\\ (x+a)^2+(y-b)^2=a^2+b^2\\ \rightarrow\boxed{O(-a,b),\hspace{4pt}R_O^2=a^2+b^2}

we have therefore obtained the center of the circle whose equation is given in the problem and the expression for the (squared) radius in terms of the parameters in the problem,

now let's address the first given fact in the problem which states that the circle is tangent to the y-axis,

Let's recall several things:

Remember that the absolute value of the x-coordinate of the center of a circle that is tangent to the y-axis, must equal the circle's radius,

(Similarly, remember that the absolute value of the y-coordinate of the center of a circle tangent to the x-axis, must equal the circle's radius).

These facts of course stem from the fact that the tangent to a circle is perpendicular to the radius at the point of tangency (meaning - to the line segment connecting the circle's center to the point of tangency).

Let's return then to the problem, we found that the center of the circle is at point:

O(a,b) O(-a,b)

and additionally from the fact that the circle is tangent to the y-axis, as mentioned before we can conclude that:

xo=RO |x_o|=R_O

we'll square both sides of this equation, then we'll substitute the x-coordinate of the circle's center and the squared radius and solve the equation we get:

xo=RO/()2xo2=RO2O(a,b),RO2=a2+b2(a)2=a2+b2a2=a2+b2b2=0b=0 |x_o|=R_O\hspace{6pt}\text{/}()^2\\ \downarrow\\ x_o^2=R_O^2\leftrightarrow \boxed{O(-a,b),\hspace{4pt}R_O^2=a^2+b^2} \\ (-a)^2=a^2+b^2\\ a^2=a^2+b^2\\ b^2=0\\ \downarrow\\ \boxed{b=0}

we have therefore found that the center of the circle whose equation is given in the problem is:

O(a,b)b=0O(a,0) O(-a,b)\stackrel{\textcolor{orange}{b=0}}{\rightarrow} O(-a,0)

meaning - the center of the circle lies on the x-axis,

however now it's worth noting that there is another important given fact:

a<0

we'll continue and use the inequality law which states that when multiplying an inequality by a negative number - the inequality sign reverses:

a<0 \hspace{6pt}\text{/}\cdot(-1)\\ -a>0

and therefore we can conclude that the center of the circle in question must be located on the positive part of the x-axis:

O(-a,0) \leftrightarrow-a>0\\ \downarrow\\ \boxed{x_o>0}

therefore the correct answer is answer d.

Answer

O lies on the positive part of the axis line

Exercise #9

Common Geometric Shapes

CO,CM C_O,\hspace{6pt}C_M

Which of them are O and M in the figure:

CO:x24x+y2+6y=12CM:x2+2x+y22y=7 C_O: x^2-4x+y^2+6y=12\\ C_M: x^2+2x+y^2-2y=7\\

How many sides do they have?


Step-by-Step Solution

In the given problem, we are asked to determine where the center of a certain circle is located in relation to the other circle,

To do this, we need to find first the characteristics of the given circles, that is - their center coordinates and their radius, let's remember first that the equation of a circle with center at point

O(xo,yo) O(x_o,y_o)

and radius R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

In addition, let's remember that we can easily determine whether a certain point is inside/outside or on a given circle by calculating the distance of the point from the center of the circle in question and comparing the result to the given circle's radius,

Let's now return to the problem and the equations of the given circles and examine them:

CO:x24x+y2+6y=12CM:x2+2x+y22y=7 C_O: x^2-4x+y^2+6y=12\\ C_M: x^2+2x+y^2-2y=7\\

Let's start with the first circle:

CO:x24x+y2+6y=12 C_O: x^2-4x+y^2+6y=12

and find its center and radius, we'll do this using the "completing the square" method,

We'll try to give this equation a form identical to the form of the circle equation, that is - we'll ensure that on the left side there will be the sum of two squared binomial expressions, one for x and one for y, we'll do this using the "completing the square" method:

To do this, first let's recall again the shortened multiplication formulas for binomial squared:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2

and we'll deal separately with the part of the equation related to x in the equation (underlined):

x24x+y2+6y=12 \underline{ x^2-4x}+y^2+6y=12

We'll continue, for convenience and clarity of discussion - we'll separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the form of the first two terms in the shortened multiplication formula (we'll choose the subtraction form of the binomial squared formula since the term with the first power 4x in the expression we're dealing with is negative):

x24x+y2+6y=12x24xc22cd+d2x22x2c22cd+d2 \underline{ x^2-4x}+y^2+6y=12 \\ \underline{ x^2-4x}\textcolor{blue}{\leftrightarrow} \underline{ c^2-2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{-2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{2}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{-2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\

It can be noticed that compared to the shortened multiplication formula (on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

{xc2d \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 2\textcolor{blue}{\leftrightarrow}d \end{cases}

Therefore, we'll identify that if we want to get from these two terms (underlined in the calculation) a binomial squared form,

we'll need to add to these two terms the term


22 2^2

However, we don't want to change the value of the expression in question, and therefore - we'll also subtract this term from the expression,

that is - we'll add and subtract the term (or expression) we need to "complete" to the binomial squared form,

In the following calculation, the "trick" is highlighted (two lines under the term we added and subtracted from the expression),

Next - we'll insert into the binomial squared form the appropriate expression (highlighted using colors) and in the last stage we'll simplify the expression further:

x22x2x22x2+2222x22x2+224(x2)24 x^2-2\cdot x\cdot 2\\ x^2-2\cdot x\cdot2\underline{\underline{+2^2-2^2}}\\ \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4\\ \downarrow\\ \boxed{ (\textcolor{red}{x}-\textcolor{green}{2})^2-4}\\

Let's summarize the development stages so far for the expression related to x, we'll do this now within the given circle equation:

CO:x24x+y2+6y=12CO:x22x2+2222+y2+6y=12CO:(x2)24+y2+6y=12 C_O: x^2-4x+y^2+6y=12 \\ C_O: \textcolor{red}{x}^2-2\cdot \textcolor{red}{x}\cdot\textcolor{green}{2}\underline{\underline{+\textcolor{green}{2}^2-2^2}}+y^2+6y=12 \\ \downarrow\\ C_O:(\textcolor{red}{x}-\textcolor{green}{2})^2-4+y^2+6y=12

We'll continue and perform an identical process for the expressions related to y in the resulting equation:

(Now we'll choose the addition form of the binomial squared formula since the term with the first power 6y in the expression we're dealing with is positive)

(x2)24+y2+6y=12(x2)24+y2+2y3=12(x2)24+y2+2y3+3232=12(x2)24+y2+2y3+329=12((x2)24+(y+3)29=12CO:(x2)2+(y+5a)2=25 (x-2)^2-4+\underline{y^2+6y}=12\\ \downarrow\\ (x-2)^2-4+\underline{y^2+2\cdot y \cdot 3}=12\\ (x-2)^2-4+\underline{y^2+2\cdot y \cdot 3\underline{\underline{+3^2-3^2}}}=12\\ \downarrow\\ (x-2)^2-4+\underline{\textcolor{red}{y}^2+2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{3}+\textcolor{green}{3}^2-9}=12\\ \downarrow\\ ( (x-2)^2-4+(\textcolor{red}{y}+\textcolor{green}{3})^2-9=12\\ C_O:\boxed{ (x-2)^2+(y+5a)^2=25}

In the last stage, we moved the free numbers to the other side and combined similar terms,

Now that we've changed the given circle equation to the form of the general circle equation mentioned earlier, we can easily extract both the center of the given circle and its radius:

(xxo)2+(yyo)2=R2CO:(x2)2+(y+3)2=25CO:(x2)2+(y(3))2=25 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ C_O:(x-\textcolor{purple}{2})^2+(y+\textcolor{orange}{3})^2=\underline{\underline{25}}\\ \downarrow\\ C_O:(x-\textcolor{purple}{2})^2+(y\stackrel{\downarrow}{- }(-\textcolor{orange}{3}))^2=\underline{\underline{25}}\\

In the last stage, we made sure to get the exact form of the general circle equation - that is, where only subtraction is performed within the squared expressions (highlighted by arrow)

Therefore we can conclude that the center of the circle is at point:


O(xo,yo)O(2,3) \boxed{O(x_o,y_o)\leftrightarrow O(2,-3)}

and extract the circle's radius by solving a simple equation:

R2=25/R=5 R^2=25\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=5}

Let's summarize the information so far:

CO:{O(2,3)R=5 C_O:\begin{cases} O(2,-3)\\ R=5 \end{cases}

Now let's approach the equation of the given second circle and find its center and radius through an identical process, here we'll do it in parallel for both variables:

CM:x2+2x+y22y=7CM:x2+2x1+y22y1=7CM:(x+1)212+(y1)212=7CM:(x+1)2+(y1)2=9CM:(x(1))2+(y1)2=32 C_M: x^2+2x+y^2-2y=7 \\ \downarrow\\ C_M: x^2+2\cdot x\cdot 1+y^2-2\cdot y\cdot 1=7 \\\\ \downarrow\\ C_M: (x+1)^2-1^2+(y-1)^2-1^2=7 \\ C_M:\boxed{ (x+1)^2+(y-1)^2=9} \\ \downarrow\\ C_M:\boxed{ (x-(1))^2+(y-1)^2=3^2} \\\\

Therefore we'll conclude that the circle's center and radius are:

CM:{M(1,1)R=3 C_M:\begin{cases} M(-1,1)\\ R=3 \end{cases}

Now in order to determine which of the options is the most correct, that is - to understand where the centers of the circles are in relation to the circles themselves, all we need to do is calculate the distance between the centers of the circles (using the distance formula between two points) and check the result in relation to the radii of the circles, let's first present the data of the two circles:

CO:{O(2,3)R=5,CM:{M(1,1)R=3 C_O:\begin{cases} O(2,-3)\\ R=5 \end{cases},\hspace{6pt} C_M:\begin{cases} M(-1,1)\\ R=3 \end{cases}

Let's remember that the distance between two points in a plane with coordinates:

A(xA,yA),B(xB,yB) A(x_A,y_A),\hspace{6pt}B(x_B,y_B)

is:

dAB=(xAxB)2+(yAyB)2 d_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}

And therefore, the distance between the centers of the circles is:

dOM=(2(1))2+(31)2dOM=9+16=25dOM=5 d_{OM}=\sqrt{(2-(-1))^2+(-3-1)^2} \\ d_{OM}=\sqrt{9+16} =\sqrt{25} \\ \boxed{d_{OM}=5}

That is, we got that the distance between the centers of the circles is 5,

Let's note that the distance between the centers of the circles


dOM d_{OM}

is equal exactly to the radius of the circle

CO C_O

That is - point M is on the circle

CO C_O

(And this follows from the definition of a circle as the set of all points in a plane that are at a distance equal to the radius of the circle from the center of the circle, therefore necessarily a point at a distance from the center of the circle equal to the radius of the circle - is on the circle)

In addition, let's note that the distance between the centers of the circles

dOM d_{OM}

is greater than the radius of the circle

CM C_M

That is - point O is outside the circle

CM C_M

Therefore, the most correct answer is answer B.

Answer

Measure angle CO C_O opposite to angle CM C_M

Exercise #10

Given Data: A(0,2) A(0,2)

And the result in the table is

O O and its explanation:

x2+8x+y24y=4 x^2+8x+y^2-4y=-4

According to the given data:


Step-by-Step Solution

In the given problem, we are asked to determine where a certain point is located in relation to a given circle,

To do this, we need to first find the characteristics of the given circle, namely- its center and radius,

Let's remember first that the equation of a circle with center at point

O(xo,yo) O(x_o,y_o)

and radius R is:

(xxo)2+(yyo)2=R2 (x-x_o)^2+(y-y_o)^2=R^2

Additionally, let's recall that we can easily determine whether a certain point is inside/outside the circle or on it, by calculating the distance of the point from the center of the circle in question and comparing the result to the given circle's radius,

Let's now return to the problem and the given circle equation and examine them:

x2+8x+y24y=4 x^2+8x+y^2-4y=-4

Let's find its center and radius, we'll do this using the "completing the square" method,

We'll try to give this equation a form identical to the general circle equation, meaning- we'll ensure that on the left side there will be a sum of two squared binomial expressions, one for x and one for y, we'll do this using the "completing the square" method:

For this, let's first recall the shortened multiplication formulas for squared binomials:

(c±d)2=c2±2cd+d2 (c\pm d)^2=c^2\pm2cd+d^2

and we'll deal separately with the part of the equation related to x in the equation (underlined):

x2+8x+y24y=4 \underline{x^2+8x}+y^2-4y=-4

We'll continue, for convenience and clarity of discussion- we'll separate these two terms from the equation and deal with them separately,

We'll present these terms in a form similar to the form of the first two terms in the shortened multiplication formula (we'll choose the addition form of the squared binomial formula since the term with the first power we're dealing with 8x has a positive sign):

x2+8x+y2+6y=12x2+8xc2+2cd+d2x2+2x4c2+2cd+d2 \underline{ x^2+8x}+y^2+6y=12 \\ \underline{ x^2+8x}\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \downarrow\\ \underline{\textcolor{red}{x}^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}} \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\

We can notice that compared to the shortened multiplication formula (on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

{xc4d \begin{cases} x\textcolor{blue}{\leftrightarrow}c\\ 4\textcolor{blue}{\leftrightarrow}d \end{cases}

Therefore, we'll identify that if we want to get from these two terms (underlined in the calculation) a squared binomial form,

we'll need to add to these two terms the term


42 4^2

However, we don't want to change the value of the expression in question, and therefore- we'll also subtract this term from the expression,

meaning- we'll add and subtract the term (or expression) needed to "complete" to a squared binomial form,

In the following calculation, the "trick" is highlighted (two lines under the term we added and subtracted from the expression) ,

Next- we'll insert into the squared binomial form the appropriate expression (highlighted with colors) and in the last stage we'll further simplify the expression:

x2+2x4x2+2x4+4242x2+2x4+4216(x+4)216 x^2+2\cdot x\cdot 4\\ x^2+2\cdot x\cdot4\underline{\underline{+4^2-4^2}}\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot \textcolor{green}{4}+\textcolor{green}{4}^2-16\\ \downarrow\\ \boxed{ (\textcolor{red}{x}+\textcolor{green}{4})^2-16}\\

Let's summarize the development stages so far for the expression related to x, we'll do this now within the given circle equation:

x2+8x+y24y=4x2+2x4+4242+y24y=4(x+4)216+y24y=4 x^2+8x+y^2-4y=-4\\ \textcolor{red}{x}^2+2\cdot \textcolor{red}{x}\cdot\textcolor{green}{4}\underline{\underline{+\textcolor{green}{4}^2-4^2}}+y^2-4y=-4 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{green}{4})^2-16+y^2-4y=-4\\

We'll continue and perform an identical process for the expressions related to y in the resulting equation:

(Now we'll choose the subtraction form of the squared binomial formula since the term with the first power we're dealing with 4y has a negative sign)

(x+4)216+y24y=4(x+4)216+y22y2=4(x+4)216+y22y2+2222=4(x+4)216+y22y2+224=4(x+4)216+(y2)24=4(x+4)2+(y2)2=16 (x+4)^2-16+\underline{y^2-4y}=-4\\ \downarrow\\ (x+4)^2-16+\underline{y^2-2\cdot y \cdot 2}=-4\\ (x+4)^2-16+\underline{y^2-2\cdot y \cdot 2\underline{\underline{+2^2-2^2}}}=-4\\ \downarrow\\ (x+4)^2-16+\underline{\textcolor{red}{y}^2-2\cdot\textcolor{red}{ y}\cdot \textcolor{green}{2}+\textcolor{green}{2}^2-4}=-4\\ \downarrow\\ (x+4)^2-16+(\textcolor{red}{y}-\textcolor{green}{2})^2-4=-4\\ \boxed{ (x+4)^2+(y-2)^2=16}

In the last stage, we moved the free numbers to the other side and combined similar terms,

Now that we've changed the given circle equation to the form of the general circle equation mentioned earlier, we can extract from the given equation both the center of the given circle and its radius simply:

(xxo)2+(yyo)2=R2CO:(x+4)2+(y2)2=16CO:(x(4))2+(y2)2=16 (x-\textcolor{purple}{x_o})^2+(y-\textcolor{orange}{y_o})^2=\underline{\underline{R^2}} \\ \updownarrow \\ C_O:(x+\textcolor{purple}{4})^2+(y-\textcolor{orange}{2})^2=\underline{\underline{16}}\\ \downarrow\\ C_O:(x-(-\textcolor{purple}{4}))^2+(y\stackrel{\downarrow}{- }\textcolor{orange}{2})^2=\underline{\underline{16}}\\

In the last stage, we made sure to get the exact form of the general circle equation - meaning- where only subtraction is performed within the squared expressions (highlighted by arrow)

Therefore we can conclude that the center of the circle is at point :


O(xo,yo)O(4,2) \boxed{O(x_o,y_o)\leftrightarrow O(-4,2)}

And extract the circle's radius by solving a simple equation:

R2=16/R=4 R^2=16\hspace{6pt}\text{/}\sqrt{\hspace{4pt}}\\ \rightarrow \boxed{R=4}

Meaning the circle's characteristics (its center and radius) are:

{O(4,2)R=4 \begin{cases} O(-4,2)\\ R=4 \end{cases}

Now in order to determine which of the options is most correct, meaning- to understand where the given point is located:

A(0,2) A(0,2)

In relation to the given circle, all we need to do is to calculate the distance between the given point and the center of the given circle (using the distance formula between two points) and check the result in relation to the circle's radius , first-

Let's remember that the distance between two points in a plane with coordinates :

A(xA,yA),B(xB,yB) A(x_A,y_A),\hspace{6pt}B(x_B,y_B)

is:

dAB=(xAxB)2+(yAyB)2 d_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}

And therefore, the distance between the given point and the center of the given circle is:

{O(4,2)A(0,2)dOA=(40)2+(22)2dOA=16+0=16dOA=4 \begin{cases} O(-4,2)\\ A(0,2) \end{cases}\\ \downarrow\\ d_{OA}=\sqrt{(-4-0)^2+(2-2)^2} \\ d_{OA}=\sqrt{16+0} =\sqrt{16} \\ \boxed{d_{OA}=4}

Meaning we got that the distance between the given point and the center of the given circle is 4,

Let's note that the distance between the given point and the center of the given circledOA d_{OA} equals exactly the circle's radius :

dOA=R=4 d_{OA}=R=4

Meaning- point A is located on the given circle,

(This follows from the definition of a circle as the set of all points in a plane that are at a distance equal to the circle's radius from the circle's center, therefore necessarily a point located at a distance from the circle's center equal to the circle's radius - is on the circle)

And therefore the most correct answer is answer c.

Answer

Table A on the given result