If the circle
$$x^2 + y^2 + 2gx + 2fy + c = 0$$
cuts the three circles
$$x^2 + y^2 – 5 = 0\space;\space x^2 + y^2 – 8x – 6y + 10 = 0
\space;\space x^2 + y^2 – 4x + 2y – 2 = 0;$$
at the extremities of their diameters, then
(A) $c = – 5 $
(B) $fg = 147/25$
(C) $g + 2f = c + 2 $
(D) $4f = 3g$
I'm not understanding a single thing. Please Help
Best Answer
You might know that the equation of the common chord of two circle whose standard equations are $S_1=0$ and $S_2=0$ , is $S_1=S_2$ .
The required circle bisects the circumference of all the other 3 circles, so its common chord with these circles should pass through their respective centres.
So, we have: $$c=-5\tag{i}$$
$$8g+6f+c=-32-18+10\tag{ii}$$
$$4g-2f+c=-8-2-2\tag{iii}$$
On solving above system of equations you can get the values of $c,f,g$.
As requested by Mick, here is a picture :
I've marked the centres and diameters of the given circles, and the required circle is shown in red.