$\text{ The distance between the origin and the tangent to the curve }$$y=e^{2x}+x^2\text{ drawn at the point }x=0\text{ is }$

$\text{ Putting }x=0,\text{ we get }y=1$

So the point is (0, 1)

$\Rightarrow \frac{dy}{dx}=2e^{2x}+2x\Rightarrow {\left(\frac{dy}{dx}\right)}_{(0,1)}=2\times 1+0=2$

$\begin{array}{r}\text{ Equation of tangent at }(0,1)\text{ is }\\ y-1=2(x-0)\Rightarrow 2x-y+1=0\end{array}$

$\begin{array}{l}\text{ Perpendicular distance from origin to the tangent }\\ =\left|\frac{1}{\sqrt{4+1}}\right|=\frac{1}{\sqrt{5}}\end{array}$