The tangent to the circle $x^2+y^2=5$ at the point $\left(1,-2\right)$ also touches the circle $x^2+y^2-8x+6y+20=0$ at the point

Equation of the tangent to $x^2+y^2=5$ at $\left(1,-2\right)$ is $x\left(1\right)+y\left(-2\right)=5.$

Let, this tangent touches $x^2+y^2-8x+6y+20=0$ at $\left(h,k\right)$
Then, $hx+ky-4\left(x+h\right)+3\left(y+k\right)+20=0$ is same as $x-2y-5=0$
$\Rightarrow \frac{h-4}{1}=\frac{k+3}{-2}=\frac{-4h+3k+20}{-5}$
$\Rightarrow 2h+k-5=0$ and $h+3k=0$
$\Rightarrow h=3,k=-1$
$\Rightarrow$ Required point is $\left(3,-1\right)$