The equation of the circle whose centre lies on $x+2y=0$ and touching $3x-4y+8=0$ and $3x-4y-28=0$ is

Given two tangents   $3x-4y+8=0-\left(1\right)$

And  $3x-4y-28=0-\left(2\right)$  are parallel

$2r=$ distance b/w the tangents

$=\frac{\left|8+28\right|}{\sqrt{9+16}}=\frac{36}{5}$

radius  $=r=\frac{18}{5}$

midway line of $\left(1\right)\times \left(2\right)$ passes through centre

midway line of $\left(1\right)\times \left(2\right)$ is $3x-4y-10=0-\left(3\right)$

and centre also lies on  $x+2y=0-\left(4\right)$

solve  $\left(3\right)\times \left(4\right)$  we get centre $=\left(2,-1\right)$

Equation of circle $25{\left(x-2\right)}^2+25{\left(y+1\right)}^2=324$