JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Parabola

The parabola $2y^2=-x$ can be written as $y^2=-\frac{x}{2}$. The equation of a tangent to this parabola is $y=mx-\frac{1}{8m}$. Since this tangent passes through $(2,0)$, we have $0=2m-\frac{1}{8m}$, which gives $16m^2=1$, so $m=\pm \frac{1}{4}$. The two tangent lines are $x-4y-2=0$ and $x+4y-2=0$. These lines are also tangent to the circle $(x-5{)}^2+y^2=r$. The distance from the center $(5,0)$ to the line $x-4y-2=0$ is $\frac{|5-0-2|}{\sqrt{1+16}}=\frac{3}{\sqrt{17}}$. This distance equals $\sqrt{r}$, so $r=\frac{9}{17}$. Therefore, $17r=9$.