JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Circle

For $x^2+y^2+2gx+2fy+c=0$ the intercepts are $2\sqrt{g^2-c}$ on the $x$-axis and $2\sqrt{f^2-c}$ on the $y$-axis. Here $2g=a$ and $2f=2a$, so

Subtracting gives $\frac{3a^2}{4}=3$, so $a^2=4$ and $a=-2$ because $a<0$, whence $c=-1$. The centre is $(1,2)$ and the radius is $\sqrt{\frac{a^2}{4}+a^2-c}=\sqrt{1+4+1}=\sqrt{6}$.
A tangent perpendicular to $x+2y=0$ has slope $2$, say $2x-y+k=0$, and its distance from the centre equals the radius:

Its distance from the origin is $\frac{|k|}{\sqrt{5}}=\frac{\sqrt{30}}{\sqrt{5}}=\sqrt{6}$, the same for either sign of $k$, so the shortest distance from the origin to such a tangent is $\sqrt{6}$.