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___

Answer 1

End points of latusrectum are $(a,\pm 2a)$ i.e. $(1,\pm 2)$. Equation of normal at $(x_1,y_1)$ is
$\frac{y-y_1}{x-x_1}=-\frac{y_1}{2a}$
i.e. $\frac{y-2}{x-1}=-\frac{2}{2}$ and $\frac{y+2}{x-1}=\frac{2}{2}$
$\Rightarrow x+y=3$ and $x-y=3$
These are tangent to $(x-3{)}^2+(y+2{)}^2=r^2$

$\therefore$ Length of perpendicular from centre of circle = Radius
$\Rightarrow \frac{|3-2-3|}{\sqrt{1^2+1^2}}=r\therefore r^2=2$