Question

If the tangent to the circle $x^2+y^2=5$ at $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$ at $P(\alpha ,\beta ),$ then the value of ${\alpha }^2+{\beta }^2$ is

Answer 1

Equation of tangent at $(1,-2)$ to the circle $x^2+y^2=5,$ is
$T=0\Rightarrow x\left(1\right)+y(-2)-5=0\Rightarrow x-2y-5=0$

Above line touches the circle $x^2+y^2-8x+6y+20=0$ at $P(\alpha ,\beta ).$
Centre of the circle is $C=(4,-3)$
$\Rightarrow P(\alpha ,\beta )$ is the foot of perpendicular from centre to the tangent $x-2y-5=0$
$\Rightarrow \frac{\alpha -4}{1}=\frac{\beta +3}{-2}=\frac{-(4+6-5)}{1^2+(-2{)}^2}\Rightarrow \alpha =3,\beta =-1\therefore {\alpha }^2+{\beta }^2=10$