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. $x^{2} \, =16 \, y+4$
• 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
$\therefore 2 \, \left(- 10 g + 0 \, \times f \, \right)= \, c+4 \, \Rightarrow - \, 20 \, g=c+4$ .......(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 $x$ & $y$ )