ExamHabit
Practise free

2000

JEE Main 2000 Mathematics question on Circles

The △PQR\triangle PQR is inscribed in the circle x2+y2=25x^2 + y^2 = 25. If QQ and RR have coordinates (3,4)(3, 4) and (−4,3)(-4, 3) respectively, then ∠QPR\angle QPR is equal to
  1. A.
    π/2\pi/2
  2. B.
    π/3\pi/3
  3. C.
    π/4\pi/4
  4. D.
    π/6\pi/6
Show answer and solution

Correct option: C

Solution

The chord QR subtends an inscribed angle ∠QPR at point P on the circle. Using the inscribed angle theorem, ∠QPR = (1/2) × (central angle subtended by QR).

The central angle equals the angle ∠QOR where O is the origin (center). The vectors OQ = (3,4) and OR = (-4,3) have dot product: (3)(-4) + (4)(3) = -12 + 12 = 0. So ∠QOR = π/2, meaning QR subtends a right angle at the center.

Therefore ∠QPR = (1/2)(π/2) = π/4.

[2000]

Practise Circles the way it is asked

A timed set of real JEE Main Circles questions, graded as the exam grades. No sign-up.

Start a Circles test

More: all JEE Main Circles questions · JEE Main 2000 · Mathematics chapter list