If the radical axis of the circles $x^2+y^2+2 g x+2 f y+c=0$ and $2 x^2+2 y^2+3 x+8 y+2 c=0$ touches the…
If the radical axis of the circles
$x^2+y^2+2 g x+2 f y+c=0$ and
$2 x^2+2 y^2+3 x+8 y+2 c=0$ touches the circle $x^2+y^2+2 x+2 y+1=0$, then
$g=\frac{3}{4}$ or $f=2$
$g \neq \frac{3}{4}, f=2$
$g=\frac{3}{4}$ or $f \neq 2$
$g=\frac{2}{5}$ or $f=1$
Solution
$
\begin{aligned}
& \text { Let point } P(a, b) \\
& \qquad \begin{array}{l}
S_1(a, b)=S_2(a, b) \\
\Rightarrow a^2+b^2+2 g a+2 f b+c=a^2+b^2+\frac{3}{2} a+4 b+c=0 \\
a\left(2 g-\frac{3}{2}\right)+b(2 f-4)=0
\end{array}
\end{aligned}
$
this is the locus of radical axis.
So, $x\left(2 g-\frac{3}{2}\right)+y(2 f-4)=0$ is radical axis of given circles.
This touched the $x^2+y^2+2 x+2 y+1=0$
So, radius $=\sqrt{1^2+1^2-1}=1$ and centre $=(-1,-1)$
So, radius of circle $=$ distance between centre and touching point.
$
1=\frac{\left|\left(\frac{3}{2}-2 g\right)+(2 f-4)\right|}{\sqrt{\left(\frac{3}{2}-2 g\right)^2+(2 f-4)^2}}
$
Taking square both sides,
$
2\left(\frac{3}{2}-2 g\right)(2 f-4)=0
$
So,
$
\begin{aligned}
\frac{3}{2}-2 g & =0 \text { or } \\
2 f-4 & =0 \\
2 g & =\frac{3}{2} \text { or } 2 f=4 \\
g & =\frac{3}{4} \text { or } f=2
\end{aligned}
$