Question

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

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

Answer 1

Answer: (C)

Let the equation of circle be
$x^2+y^2+2gx+2fy+c=0$
where, centre $(-g,-f)$
The centre of given circle $x^2+y^2-20x+4=0$ is $(10,0)$
Condition of two circles cut.
$\therefore 2\left(g_1{g}_2+f_1{f}_2\right)=c_1+c_2$
$2(-g\times 10+0\times (-f)=c+4$
$2(-10g)=c+4$
Also, circle touch the line $x=2$.
$\therefore$ The perpendicular distance from centre to the circle is equal to radius of the circle.
$\therefore \frac{|-g-2|}{\sqrt{1}}=\sqrt{g^2+f^2-c}$
$\Rightarrow (g+2)=\sqrt{g^2+f^2-c}$
$\Rightarrow g^2+4+4g=g^2+f^2-c$
$\Rightarrow f^2-4g-c-4=0$
$\Rightarrow f^2-4g+4+20g-4=0$
$\Rightarrow f^2+16g=0$
Hence, the locus of $(-g,-f)$ is
$y^2-16x=0$
$\Rightarrow y^2=16x$