Question

$P$ is a point on the parabola $y^2=16x$, where abscissa and ordinate are equal. Equation of a circle passing through the focus and touching the parabola at $P$ can be

• A. $x^2+y^2+52x+48y-160=0$
• B. $x^2+y^2-4x=0$
• C. $x^2+y^2+4x=0$
• D. $x^2+y^2-52x+8y+192=0$

Answer 1

Answer: (B), (D)

The correct options are
B $x^2+y^2-4x=0$
D $x^2+y^2-52x+8y+192=0$

Given equation of parabola is $y^2=16x$
Focus is at $(4,0)$
Coordinates of point $P$ are $(0,0)$ and $(16,16)$
Equation of tangents at $P$ are
$2y-x-16=0$ and $x=0$
Required equation of circle touching at $(16,16)$
$(x-16{)}^2+(y-16{)}^2+\lambda (2y-x-16)=0$
Since, it passes through $(4,0)$
$\lambda =20$
$\Rightarrow (x-16{)}^2+(y-16{)}^2+20(2y-x-16)=0$
$x^2+y^2-32x-32y+512+40y-20x-320=0$
$x^2+y^2-52x+8y+192=0$
Required equation of circle touching at $(0,0)$
$(x-0{)}^2+(y-0{)}^2+\lambda \left(x\right)=0$
Since, it passes through $(4,0)$
$\lambda =-4$
$x^2+y^2-4x=0$