JEE Main 2017 — Circle Question with Solution

Let $y=mx+c$ be the common tangent.

Then $c^2=4\left(1+m^2\right) ...\left(1\right)$ (condition of tangency for circle)

Solving with $x^2=4y$, we get $x^2=4\left(mx+c\right)$

i.e. $x^2-4mx-4c=0 ...\left(1\right)$

Being a tangent, $\left(1\right)$ must have equal roots.

$\Rightarrow {\left(-4m\right)}^2=4\left(1\right)\left(-4c\right)\Rightarrow m^2=-c ...\left(2\right)$

From $\left(1\right) \& \left(2\right), c^2=4-4c \& c<0$

$\Rightarrow c^2+4c-4=0\Rightarrow c=-2\sqrt{2}-2$ (As $c<0$, so $c\ne 2\sqrt{2}-2$).

So, both tangents have common $y$ intercept and thus intersect at $P\left(0,-2-\sqrt{2}\right)$.

Thus, distance of $P$ from origin is $2\left(\sqrt{2}+1\right)$ units.