Question

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

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

Answer 1

The correct option is D $y^2=16x$

Let circle be

$x^2+y^2+29x+26y+c=0$

with center $(-2,-6)$

cuts orthognally $x^2+y^2-2x+4=0$

$2(-10)g+0=4+c$

$-20g=4+c---\left(1\right)$

Radius $=\sqrt{9^2+6^2+4+20g}$

$\perp$ distance from $(-9,-6)$ on liner $x=2$ is Radius

$-9-2=\sqrt{9^2+6^2+4+20g}$

$9^2+4+4g=9^2+6^2+4+20g$

$6^2=-16g$

Put $-9=x-6=y$

$y^2=16x$

$D$ is correct