Circumcircle of Triangle

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,

$$\boxed{\frac{\sqrt{10}}{2}}$$