Question

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

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

Answer 1

Answer: (C)

Let the equation of circle be
$x^{2}+y^{2}+2\, g x+2\, f \,y+c=0$
where, centre $(-g,-f)$
The centre of given circle $x^{2}+y^{2}-20 x+4=0$ is $(10,0)$
Condition of two circles cut.
$\therefore 2\left(g_{1} \,g_{2}+f_{1}\, f_{2}\right) =c_{1}+c_{2} $
$2(-g \times 10+0 \times(-f)=c+4 $
$2(-10 \,g) =c+4$
Also, circle touch the line $x=2$.
$\therefore $ The perpendicular distance from centre to the circle is equal to radius of the circle.
$\therefore \frac{|-g-2|}{\sqrt{1}}=\sqrt{g^{2}+f^{2}-c}$
$\Rightarrow (g+2)=\sqrt{g^{2}+f^{2}-c}$
$\Rightarrow g^{2}+4+4 \,g=g^{2}+f^{2}-c$
$\Rightarrow f^{2}-4\, g-c-4=0$
$\Rightarrow f^{2}-4 \,g+4+20\, g-4=0$
$\Rightarrow f^{2}+16\, g=0$
Hence, the locus of $(-g,-f)$ is
$y^{2}-16 x=0$
$ \Rightarrow y^{2}=16 x$