Circle Equation Analysis: x²+2ax+y²-2by=0 with Negative Parameter a

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

$O(x_o,y_o)$

$$and radius$R$ is:

$(x-x_o)^2+(y-y_o)^2=R^2$

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

$x^2+2ax+y^2-2by=0$

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

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

$x^2+2ax+y^2-2yb=0 \\ \downarrow\\ x^2+2\cdot x\cdot \textcolor{blue}{a}+y^2-2\cdot y\cdot \textcolor{red}{b}=0 \\ \downarrow\\ (x-\textcolor{blue}{a})^2-\textcolor{blue}{a}^2+(y-\textcolor{red}{b})^2-\textcolor{red}{b}^2=0\\ (x+a)^2+(y-b)^2=a^2+b^2\\ \rightarrow\boxed{O(-a,b),\hspace{4pt}R_O^2=a^2+b^2}$

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:

$O(-a,b)$

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

$|x_o|=R_O$

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:

$|x_o|=R_O\hspace{6pt}\text{/}()^2\\ \downarrow\\ x_o^2=R_O^2\leftrightarrow \boxed{O(-a,b),\hspace{4pt}R_O^2=a^2+b^2} \\ (-a)^2=a^2+b^2\\ a^2=a^2+b^2\\ b^2=0\\ \downarrow\\ \boxed{b=0}$

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

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

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.