For the parabola $$y=x^2$$ the slope of the tangent is $$\frac{dy}{dx}=2x$$
Given,
$$2x=4$$
$$x=2$$
Hence,
$$P=(2,4)$$
For the circle $$x^2+y^2=2$$ differentiating, $$2x+2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{x}{y}$$
Given,
$$-\frac{x}{y}=-1$$
$$x=y$$
Since $$Q$$ lies on $$x^2+y^2=2$$
we get $$2x^2=2$$
$$x=y=1$$
Hence,
$$Q=(1,1)$$
For the ellipse $$x^2+4y^2=8$$ differentiating, $$2x+8y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{x}{4y}$$
Given,
$$-\frac{x}{4y}=-\frac12$$
$$x=2y$$
Substituting into $$x^2+4y^2=8$$ gives $$4y^2+4y^2=8$$
$$8y^2=8$$
$$y=1,\quad x=2$$
Hence, $$R=(2,1)$$
Now, $$PR=|4-1|=3$$ and $$QR=|2-1|=1$$
Since $$PR$$ is vertical and $$QR$$ is horizontal,
$$\angle PRQ=90^\circ$$
Therefore, the circumradius of triangle $$PQR$$ is half the hypotenuse.
Now,
$$PQ=\sqrt{(2-1)^2+(4-1)^2}$$
$$=\sqrt{10}$$
Hence,
$$\text{Radius}=\frac{PQ}{2}$$
$$=\frac{\sqrt{10}}{2}$$
Therefore,
$$\boxed{\frac{\sqrt{10}}{2}}$$
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