Let the circle $S \equiv x^2+y^2+2 g x+2 f y+c=0$ cut the circles $x^2+y^2-2 x+2 y-2=0$ and $x^2+y^2+4 x-6…
Let the circle $S \equiv x^2+y^2+2 g x+2 f y+c=0$ cut the circles $x^2+y^2-2 x+2 y-2=0$ and $x^2+y^2+4 x-6 y+9=0$ orthogonally. If the centre of the circle $S=0$ lies on the line $2 \mathrm{x}+3 \mathrm{y}-2=0$, then $2 \mathrm{~g}+\mathrm{f}=$
$\mathrm{c}$
$c+f$
$2 \mathrm{~g}-\mathrm{c}$
$c-f$
Solution
In circle $S=x^2+y^2+2 g x+2 f y+c=0...(1)$ centre $=(-g,-f), g_{\hat{i}}=g, f_1=f, c_1=c$
In circle $x^2+y^2-2 x+2 y-2=0...(2)$
$
g_2=-1, f_2=1, c_2=-2
$
In circle $x^2+y^2+4 x-6 y+9=0...(3)$
$
g_3=2, f_3=-3, c_3=9
$
Since eq (1) \& (2) are orthogonal hence
$
\begin{aligned}
& \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\
& \Rightarrow-2 g+2 f=c-2...(4)
\end{aligned}
$
Since eq (1) \& (3) are also orthogonal
Hence $4 g-6 f=c+9...(5)$
Since center $(-g,-f)$ lies on $2 x+3 y-2=0$ Hence
$
\Rightarrow-2 g-3 f-2=0
$
Solving eq (4), (5) and (6), we get
$
g=\frac{1}{2}, f=-1, c=-1
$
Therefore $2 g+f=0=c-f$