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: Understanding the Concept:
We need to find the specific coordinates of points \( P, Q, \) and \( R \) based on the slope conditions provided for each conic section. Once the coordinates are found, we can identify the geometric relationship between them. Often in such problems, the points form a right-angled triangle, which simplifies finding the circumradius significantly.
Step 2: Key Formula or Approach:
For any curve \( f(x, y) = 0 \), the slope of the tangent is \( m = \frac{dy}{dx} \).
Circumradius of a right triangle is half of its hypotenuse.
Distance formula: \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
Step 3: Detailed Explanation:
1. Point \( P \):
Parabola: \( y = x^2 \implies \frac{dy}{dx} = 2x \).
Given slope \( m = 4 \implies 2x = 4 \implies x = 2 \).
Then \( y = 2^2 = 4 \).
So, \( P = (2, 4) \).
2. Point \( Q \):
Circle: \( x^2 + y^2 = 2 \). Differentiating implicitly: \( 2x + 2y y' = 0 \implies y' = -x/y \).
Given slope \( m = -1 \implies -x/y = -1 \implies x = y \).
Substitute into circle: \( x^2 + x^2 = 2 \implies 2x^2 = 2 \implies x = 1 \) (since \( Q \) is in 1st quadrant).
So, \( Q = (1, 1) \).
3. Point \( R \):
Ellipse: \( x^2 + 4y^2 = 8 \). Differentiating implicitly: \( 2x + 8y y' = 0 \implies y' = -x/4y \).
Given slope \( m = -1/2 \implies -x/4y = -1/2 \implies 2x = 4y \implies x = 2y \).
Substitute into ellipse: \( (2y)^2 + 4y^2 = 8 \implies 4y^2 + 4y^2 = 8 \implies 8y^2 = 8 \implies y = 1 \).
Then \( x = 2(1) = 2 \).
So, \( R = (2, 1) \).
4. Geometry of Triangle \( PQR \):
Coordinates: \( P(2, 4), Q(1, 1), R(2, 1) \).
Observe that \( P \) and \( R \) share the same x-coordinate (\( x=2 \)), meaning the side \( PR \) is vertical.
Observe that \( Q \) and \( R \) share the same y-coordinate (\( y=1 \)), meaning the side \( QR \) is horizontal.
Thus, \( \triangle PQR \) is a right-angled triangle at vertex \( R(2, 1) \).
5. Calculation of Radius:
The hypotenuse of this right triangle is \( PQ \).
Length of \( PQ = \sqrt{(2-1)^2 + (4-1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10} \).
The radius of the circumcircle \( r = \frac{1}{2} \text{hypotenuse} = \frac{\sqrt{10}}{2} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}} \).
Step 4: Final Answer:
The radius is \( \sqrt{5/2} \).

Answer 2

Concept:

• The circumcenter is the point equidistant from all three vertices. Find it by intersecting the perpendicular bisectors of two sides, then the circumradius is just its distance to any one vertex.

Step 1: Find points $P$, $Q$, $R$ exactly as before.
$P=(2,4)$ from the parabola, $Q=(1,1)$ from the circle, $R=(2,1)$ from the ellipse.

Step 2: Find the perpendicular bisector of $QR$.
$Q=(1,1)$, $R=(2,1)$ — this side is horizontal, so its perpendicular bisector is the vertical line through the midpoint $(1.5,1)$:
$x = 1.5$

Step 3: Find the perpendicular bisector of $PR$.
$P=(2,4)$, $R=(2,1)$ — this side is vertical, so its perpendicular bisector is the horizontal line through the midpoint $(2, 2.5)$:
$y = 2.5$

Step 4: Intersect the two bisectors to get the circumcenter.
Circumcenter $= (1.5,\ 2.5)$

Step 5: Compute the distance from the circumcenter to any vertex, say $R=(2,1)$.
$R_{\text{circle}} = \sqrt{(2-1.5)^2+(1-2.5)^2} = \sqrt{0.25+2.25} = \sqrt{2.5} = \sqrt{\dfrac52}$

Final Answer: $\sqrt{\dfrac52}$