Solve Circle Equation x²-10ax+y²-4yb=-20ab with Parameter Conditions

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

$O(x_o,y_o)$

$$and its 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:

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

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

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

$x^2-10ax+y^2-4yb=-20ab \\ \downarrow\\ x^2-2\cdot x\cdot \textcolor{blue}{5a}+y^2-2\cdot y\cdot \textcolor{red}{2b}=-20ab \\ \downarrow\\ (x-\textcolor{blue}{5a})^2-(\textcolor{blue}{5a})^2+(y-\textcolor{red}{2b})^2-(\textcolor{red}{2b})^2=-20ab\\ (x-5a)^2+(y-2b)^2=(5a)^2-20ab+(2b)^2\\ \rightarrow\boxed{O(5a,2b),\hspace{4pt}R_O^2=(5a)^2-20ab+(2b)^2}$

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):

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

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

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

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

and note that although this expression: $(5a-2b)^2$ 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:

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

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:

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

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: $y=x$then it must satisfy it, meaning it must be true that:

$\textcolor{red}{\boxed{y_o=x_o}}$

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 parameters$a,\hspace{2pt}b$for which the circle's center (represented by the given equation) lies on the line: $y=x$.

Therefore- the correct answer is answer C.