Question

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

• A. $y^2=16x+4$
• B. $x^2=16y$
• C. $x^2=16y+4$
• D. $y^2=16x$

Answer 1

Answer: (D)

Let the general equation of circle be
$x^2+y^2+2gx+2fy+c=0$ ....... (i)
It cuts the circle $x^2+y^2-20x+4=0$ orthogonally
$\therefore 2\left(-10g+0\times f\right)=c+4\Rightarrow -20g=c+4$ .......(ii)
$\because$ circle (i) touches $x=2$
therefore, perpendicular distance from centre to the tangent to the circle=radius
$\Rightarrow \left|\frac{-g+0-2}{\sqrt{1^2+0^2}}\right|=\sqrt{g^2+f^2-c}$
$\Rightarrow {\left(g+2\right)}^2=g^2+f^2-c$
$\Rightarrow g^2+4+4g=g^2+f^2-c\Rightarrow 4g+4=f^2-c$ ... (iii)
on eliminating $c$ from (ii) and (iii) we get
$-16g+4=f^2+4\Rightarrow f^2+16g=0$
Hence, locus of $\left(-g,-f\right)$ is,
$y^2-16x=0$ (replacing $-f$ & $-g$ by $x$ & $y$ )