Question

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 \(-\frac12\). Then the radius of the circle passing through the points \(P\), \(Q\) and \(R\) is:

• A. \(\sqrt{10}\)
• B. \(\sqrt5\)
• C. \(\sqrt{\frac52}\)
• D. \(2\sqrt5\)

Hint:

Find the three points P, Q and R one at a time using the given slope at each point, this only needs basic differentiation on the parabola, the circle and the ellipse in turn. Once you have all three coordinates, look closely at the three given slopes, 4, -1 and -1 over 2, and check whether any two of the three connecting lines have a special relationship with each other before jumping into the general circumcircle formula. Checking the slopes first can save a lot of algebra compared to writing out the full equation of a circle through three points.

Answer 1

The Correct Option is C

Step 1: Find point \(P\).
For parabola: \[ y=x^2 \] Slope of tangent: \[ \frac{dy}{dx}=2x \] Given: \[ 2x=4 \] \[ x=2 \] Thus: \[ y=2^2=4 \] Hence: \[ P=(2,4) \]

Step 2: Find point \(Q\).
Given circle: \[ x^2+y^2=2 \] Differentiating: \[ 2x+2y\frac{dy}{dx}=0 \] \[ \frac{dy}{dx}=-\frac{x}{y} \] Given slope: \[ -\frac{x}{y}=-1 \] \[ x=y \] Substituting in: \[ x^2+y^2=2 \] \[ 2x^2=2 \] \[ x=1 \] Since first quadrant: \[ y=1 \] Thus: \[ Q=(1,1) \]

Step 3: Find point \(R\).
Given 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: \[ (2y)^2+4y^2=8 \] \[ 4y^2+4y^2=8 \] \[ 8y^2=8 \] \[ y=1 \] Thus: \[ x=2 \] Hence: \[ R=(2,1) \]

Step 4: Find the circumradius of triangle \(PQR\).
Coordinates: \[ P=(2,4),\quad Q=(1,1),\quad R=(2,1) \] Side lengths: \[ QR=1 \] \[ PR=3 \] \[ PQ=\sqrt{(2-1)^2+(4-1)^2} \] \[ =\sqrt{10} \] Triangle \(PQR\) is right angled at \(R\). Circumradius of right triangle: \[ =\frac{\text{Hypotenuse}}{2} \] Thus: \[ \text{Radius}=\frac{\sqrt{10}}{2} \] \[ =\sqrt{\frac52} \]

Step 5: Identify the correct option.
Therefore: \[ \boxed{\mathrm{(C)\ }\sqrt{\frac52}} \]

Answer 2

Concept:

• Once P, Q and R are known, the circumradius of any triangle can be found from its three side lengths and its area using $R = \dfrac{abc}{4\cdot\text{Area}}$, without needing to notice any special angle first.

Step 1: Find point $P$ on the parabola.
$y=x^2 \Rightarrow \dfrac{dy}{dx}=2x=4 \Rightarrow x=2,\ y=4$. So $P=(2,4)$.

Step 2: Find point $Q$ on the circle.
$x^2+y^2=2 \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}=-1 \Rightarrow x=y$. Substituting: $2x^2=2 \Rightarrow x=1,\ y=1$ (first quadrant). So $Q=(1,1)$.

Step 3: Find point $R$ on the ellipse.
$x^2+4y^2=8 \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{4y}=-\dfrac12 \Rightarrow x=2y$. Substituting: $8y^2=8 \Rightarrow y=1,\ x=2$. So $R=(2,1)$.

Step 4: Find the three side lengths.
$PQ=\sqrt{(2-1)^2+(4-1)^2}=\sqrt{10}$, $\ QR=\sqrt{(2-1)^2+(1-1)^2}=1$, $\ PR=\sqrt{(2-2)^2+(4-1)^2}=3$

Step 5: Find the area of triangle PQR using the coordinate (shoelace) formula.
$\text{Area}=\dfrac12\left|x_P(y_Q-y_R)+x_Q(y_R-y_P)+x_R(y_P-y_Q)\right|$
$=\dfrac12\left|2(1-1)+1(1-4)+2(4-1)\right|=\dfrac12|0-3+6|=1.5$

Step 6: Apply the circumradius formula.
$R=\dfrac{PQ\cdot QR\cdot PR}{4\cdot\text{Area}}=\dfrac{\sqrt{10}\times1\times3}{4\times1.5}=\dfrac{3\sqrt{10}}{6}=\dfrac{\sqrt{10}}{2}=\sqrt{\dfrac52}$

Final Answer: $\sqrt{\dfrac52}$