Let $P$ be the point on the parabola $y = x^2$ such that the slope of the tangent to the parabola at the point $P$ is $4$. Let $Q$ be the point in the first quadrant lying on the circle $x^2 + y^2 = 2$ such that the slope of the tangent to the circle at the point $Q$ is $-1$. Let $R$ be the point in the first quadrant lying on the ellipse $x^2 + 4y^2 = 8$ such that the slope of the tangent to the ellipse at the point $R$ is $-\dfrac{1}{2}$.
Then the radius of the circle passing through the points $P$, $Q$, and $R$ is
Step-by-Step Solution
Key Concept: To find the circumradius use $R = \dfrac{abc}{4K}$ where $a, b, c$ are the side lengths and $K$ is the area. A right angle in the triangle makes the hypotenuse the diameter, simplifying the calculation.
**Step 1: Find P on the parabola**
$y = x^2 \Rightarrow y' = 2x = 4 \Rightarrow x = 2,\; y = 4$. So $P = (2, 4)$.
**Step 2: Find Q on the circle**
Implicit differentiation of $x^2+y^2=2$: $y' = -x/y = -1 \Rightarrow x = y$. Substituting: $2x^2=2 \Rightarrow x=1$. So $Q = (1, 1)$.
**Step 3: Find R on the ellipse**
Differentiating $x^2+4y^2=8$: $y' = -x/(4y) = -1/2 \Rightarrow x = 2y$. Substituting: $4y^2+4y^2=8 \Rightarrow y=1, x=2$. So $R = (2, 1)$.
**Step 4: Compute circumradius of $\triangle PQR$**
$PQ = \sqrt{(2-1)^2+(4-1)^2} = \sqrt{10}$, $QR = \sqrt{(2-1)^2+(1-1)^2} = 1$, $PR = \sqrt{0+(4-1)^2} = 3$. Area $= \tfrac{1}{2} \cdot QR \cdot PR = \tfrac{3}{2}$ (right angle at $R$ since $QR \perp PR$). Circumradius $= \dfrac{PQ \cdot QR \cdot PR}{4 \cdot \text{Area}} = \dfrac{\sqrt{10} \cdot 1 \cdot 3}{4 \cdot \frac{3}{2}} = \dfrac{\sqrt{10}}{2}$.
Correct Answer: C