Question

If the normals of the parabola $y^2=4x$ drawn at the end points of its latus rectum are tangents to the circle $(x-3{)}^2+(y+2{)}^2=r^2$, then the value of $r^2$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A

$2$

Slope of tangents at the end points of the latus rectum of $y^2=4x$ are $\pm 1$.
Hence, the slopes of normals are $\pm 1$.
Equations of normals are $y-2=-1(x-1)$ and $y+2=1(x-1)$
$x+y-3=0$ and $x-y-3=0$.
If these are tangents to the circle $(x-3{)}^2+(y+2{)}^2=r^2$,
The distance of the centre of the circle $(3,-2)$ from the tangents is equal to ${}^{\prime }{r}^{\prime }$.
$\left|\frac{3+(-2)-3}{\sqrt{1^2+1^2}}\right|=r\Rightarrow r=\sqrt{2}\Rightarrow r^2=2$