Question 2

Step 1: Find point $P$. For $y = x^2$, $\frac{dy}{dx} = 2x$. Given slope is $4$, so $2x = 4 \implies x = 2$. Then $y = 2^2 = 4$. So $P = (2, 4)$.

Step 2: Find point $Q$. For $x^2 + y^2 = 2$, $2x + 2yy' = 0 \implies y' = -x/y$. Given slope is $-1$, so $-x/y = -1 \implies x = y$.

Substituting into circle equation: $x^2 + x^2 = 2 \implies 2x^2 = 2 \implies x = 1$ (first quadrant). So $Q = (1, 1)$.

Step 3: Find point $R$. For $x^2 + 4y^2 = 8$, $2x + 8yy' = 0 \implies y' = -x/4y$. Given slope is $-1/2$, so $-x/4y = -1/2 \implies x = 2y$.

Substituting into ellipse equation: $(2y)^2 + 4y^2 = 8 \implies 8y^2 = 8 \implies y = 1$ (first quadrant). Then $x = 2(1) = 2$. So $R = (2, 1)$.

Step 4: Find the radius of circle through $P(2, 4), Q(1, 1), R(2, 1)$.

Note that $P$ and $R$ have the same $x$-coordinate, and $Q$ and $R$ have the same $y$-coordinate. Thus $\triangle PQR$ is a right-angled triangle at $R$.

The hypotenuse is $PQ = \sqrt{(2-1)^2 + (4-1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.

The radius of the circumcircle is half the hypotenuse: Radius $= \frac{\sqrt{10}}{2} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}}$.