Question

The locus of centre of the circle which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and also touches the line $x=2$ is $y^2=ax.$ Then $a$ is

Answer 1

Let $x^2+y^2+2gx+2fy+c=0\dots \left(1\right)$ be the circle.
Given that circles cut each other orthogonally.
$\therefore 2f_1{f}_2+2g_1{g}_2=c_1+c_2$
$\Rightarrow 2g(-10)+2f\left(0\right)=c+4$
$\Rightarrow -20g=c+4\cdots \left(2\right)$

Circle $\left(1\right)$ touches the line $x-2=0$
$\Rightarrow \sqrt{g^2+f^2-c}=\left|\frac{-g-2}{\sqrt{1^2}}\right|$
$\Rightarrow g^2+f^2-c=(g+2{)}^2$
$\Rightarrow f^2-c=4g+4$
$\Rightarrow f^2-4g=c+4\dots \left(3\right)$

From $\left(2\right)$ and $\left(3\right),$
$f^2-4g=-20g$
$\Rightarrow f^2=-16g$
$\Rightarrow (-f{)}^2=16(-g)$
$\therefore$ Locus of $(-g,-f)$ is $y^2=16x$
$\Rightarrow a=+16$