Question

If the tangent drawn at $(1,2)$ to the circle $x^2+y^2-4x+2y-5=0$ is normal to the circle $x^2+y^2-4x+ky-1=0$, then the value of $|3k|$ is

Answer 1

Given circle is $x^2+y^2-4x+2y-5=0$
Equation of tangent at $(1,2)$ is
$x{x}_1+y{y}_1-2(x+x_1)+(y+y_1)-5=0\Rightarrow x+2y-2(x+1)+(y+2)-5=0\Rightarrow x-3y+5=0$

As it is normal to the circle $x^2+y^2-4x+ky-1=0$, so it passes through the centre of circle i.e. $(2,-\frac{k}{2})$
$\Rightarrow 2+\frac{3k}{2}+5=0$
$\Rightarrow k=-\frac{14}{3}$
$\therefore |3k|=14$