Question

The radius of the circle whose two normals are represented by the equation $x^{2}-5 x y-5 x+25 y=0$ and which touches externally the circle $x^{2}+y^{2}-2 x+4 y-4=0$ is equal to________.

Answer 1

Normals are
$x^{2}-5 x y-5 x+25 y=0$
or $(x-5)(x-5 y)=0$
Hence centre of first circle is $(5,1)$.
Given circle is
or $ (x-1)^{2}+(y+2)^{2}=3^{2}$
Centre of this circle is $(1,-2)$ and radius is $3 .$
Since circles touch externally,
$ C_{1} C_{2}=r_{1}+r_{2} $
$\Rightarrow 5=3+r_{2} $
$\therefore r_{2}=2$