Q) Suppose the circle with equation $x^{2} + y^{2} + 2fx + 2gy + c = 0$ cuts the parabola $y^{2} = 4ax,(a> 0)$ at four distinct points. If $d$ denotes the sum of the ordinates of these four points, then the set of possible values of $d$ is
(A) $\{ 0 \}$
(B) $(-4a,4a)$
(C) $(-a,a)$
(D) $(-\infty ,\infty )$
My Approach :- Since , the equation of circle is $x^{2} + y^{2} + 2fx + 2gy + c = 0$ . I can write it as $(x-(-f))^{2} + (y-(-g))^{2} = \left ( \sqrt{f^{2}+ g^{2}-c} \right )^{2}$
So, center of this circle is $(-f,-g)$ and radius is $ \sqrt{f^{2}+ g^{2}-c} $. Now, I assumed $4$ distinct points where given circle and parabola are intersecting are $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3}),(x_{4},y_{4})$.
I have to find here $d=y_{1}+y_{2}+y_{3}+y_{4}.$
Now these above $4$ points should be at the same distance from the center of a circle which is equal to radius.
So, $(y_{1}+g)^{2} + (x_{1}+f)^{2} = f^{2}+g^{2}-c$ and point $(x_{1},y_{1})$ satisfy the equation of given parabola also. So, $y_{1}^{2}=4ax_{1}$ (or) $x_{1}=\frac{y_{1}^{2}}{4a}$. So, equation becomes $(y_{1}+g)^{2} + (\frac{y_{1}^{2}}{4a}+f)^{2} = f^{2}+g^{2}-c$
Now,I stuck here because this equation contains terms like $y_{1}^{2}$, $y_{1}$, $y_{1}^{4}$. If I let $y_{1}^{2}=z$ then it also not help to easily solve this equation for the value of $y_{1}$.
I also tried to take some values of $a,f,g$ to eliminate options but got stuck again.
Please help me to tell the procedure to solve this question.
Hint
You have$$x^{2} + y^{2} + 2fx + 2gy + c = 0\qquad \text{and} \qquad y^2=4ax$$
Replace $x$ by $\frac{y^2}{4a}$ to get $$y^4+ 8a\left(2a+ f\right)\,y^2+32 a^2 g \,y+16 a^2 c=0$$
What is the sum of the roots ? No case to study !