Let P be the point on the parabola y=x2 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 x2+y2=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 x2+4y2=8 such that the slope of the tangent to the ellipse at the point R is −21. Then the radius of the circle passing through the points P,Q and R is
(A)10
(B)5
(C)25✓ Correct
(D)25
Answer: (C)
Solution
For parabola y=x2: dxdy=2x=4 gives x=2, so P(2,4).
For circle x2+y2=2: dxdy=−yx=−1 gives x=y. In first quadrant, x2+x2=2 gives x=1, so Q(1,1).
For ellipse x2+4y2=8: dxdy=−4yx=−21 gives x=2y. In first quadrant, (2y)2+4y2=8 gives y=1, x=2, so R(2,1).
The points are P(2,4), Q(1,1), R(2,1).
Note that QP=(1,3) and QR=(1,0). We have QP⋅QR=1, ∣QP∣=10, ∣QR∣=1.
Since PR=(0,−3) is perpendicular to QR=(1,0), the triangle PQR has a right angle at R.
For a right triangle, the hypotenuse is the diameter of the circumcircle. Here ∣PQ∣=10, so radius =210=25.
JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option
