Question

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

• A. $y^2=4(4x+1)$
• B. $y^2=16x$
• C. $x^2=4(4y+1)$
• D. $y^2=-16x$

Answer 1

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

Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$
As this cuts $x^2+y^2-20x+4=0$ orthogonally, so
$2(-10g)=c+4\cdots \left(1\right)$

$x-2=0$ is tangent to circle, so distance from centre to the line is equal to radius,
$\frac{|-g-2|}{1}=r\Rightarrow \frac{|g+2|}{1}=\sqrt{g^2+f^2-c}\Rightarrow (g+2{)}^2=g^2+f^2-c$
Using equation $\left(1\right)$, we get
$4g+4=f^2+20g+4\Rightarrow f^2+16g=0$
$\Rightarrow (-f{)}^2-16(-g)=0$

Hence, the locus of the centre is $y^2=16x$