Circle bisecting circumference of another circle

Question

The circle $x^2 + y^2 +2gx +2fy +c=0$ bisects the circumference of the circle $x^2 + y^2 +2ax +2by +d=0$. Then:

A) $2a(g-a) +2b(f-b)=c-d$

B) $2g(g-a) +2f(f-b)=d-c$

C) $2a(g+a) +2b(f+b)=c+d$

D) $2g(g+a) +2f(f+b)=c+d$

Answer 1

Consider two circles $$C_1\equiv x^2 + y^2 +2gx +2fy +c=0$$ $$\&$$ $$C_2\equiv x^2 + y^2 +2ax +2by +d=0$$ as shown in the figure below

As the common chord passes through $(x_1,y_1)\; \& \;(x_2,y_2)$, so it will satisfy both the circles $C_1$ and $C_2$, so on solving both the equations that you get when you put both the point of intersections into the equation of both the circles, you will end up with

$$2(g-a)x_1 + 2 (f-b)y_1 + c-d =0\tag{1}$$ $$2(g-a)x_2 + 2 (f-b)y_2 + c-d =0\tag{2}$$

From $(1)$ and $(2)$, we can conclude that the equation of common chord is $$\ell_{cc}\equiv 2(g-a)x+2(f-b)y+c-d =0$$

Now, lets turn to your question, since $C_1$ bisects $C_2$ (which means that the circumference of the circle is divided into two equal parts), so the common chord of both the circles should be the diameter of $C_2$. As, the diameter is a chord that passes through the center of the circle, hence $\ell_{cc}$ also passes through the center of $C_2$, i.e. $(a,b)$.

So, the answer is $$\bbox[5px, border:2px solid #C0A000]{2(g-a)a+2(f-b)b=d-c}$$