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)

$y^{2}=16 x$

Answer 2

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 $y^{2}-16 x=0$ $\Rightarrow y^{2}=16 x$