JEE Main 2013 — Circle Question with Solution

Tangent to $y^2=4\sqrt{5}x$ will be $y=mx+\frac{\sqrt{5}}{m}$ 
As tangent to $x^2+y^2=\frac{5}{2}$ will be $y=mx\pm \frac{\sqrt{5}}{\sqrt{2}}\sqrt{1+m^2}$
Hence for common tangent

$\frac{\sqrt{5}}{m}=\frac{\sqrt{5}}{\sqrt{2}}\sqrt{1+m^2}$
$\Rightarrow m^4+m^2-2=0$

$\Rightarrow m^2=1$ , $m^2=-2$ (Not Possible)

$\Rightarrow m=\pm 1$
Hence common tangent is $y=x+\sqrt{5}$

Both the statements are correct

The values of $m$ also satisfy the equation in statement - $\mathrm{II}$ but it is not a correct explanation for statement - $\mathrm{I}$.