JEE Main 2021 · 26 August · evening · Single correct · +4 / −1

A circle C touches the line x = 2y at the point (2, 1) and intersects the circle
C₁ : x² + y² + 2y - 5 = 0 at two points P and Q such that PQ is a diameter of C₁. Then the diameter of C is :

  1. (a) 7√5
  2. (b) 15
  3. (c) √285
  4. (d) 4√15
Answer and solution

Answer (a) 7√5

Solution

Maths sir · 4 steps

Use the family of circles tangent at (2, 1), then force the common chord through the centre of C₁.

  1. Family of circles. Every circle touching the line x - 2y = 0 at (2, 1) has the form (x-2)² + (y-1)² + λ(x - 2y) = 0. Expanding gives x² + y² + (λ - 4)x - (2λ + 2)y + 5 = 0.
  2. Common chord. Subtract C₁ : x² + y² + 2y - 5 = 0 from this. What is left is the line through P and Q: (λ - 4)x - (2λ + 4)y + 10 = 0.
  3. Use the diameter. PQ is a diameter of C₁, so the chord passes through its centre (0, -1). That gives (2λ + 4) + 10 = 0, so λ = -7 and C is x² + y² - 11x + 12y + 5 = 0.

    r² = (11/2)² + 6² - 5 = 121/4 + 31 = 245/4

  4. So. The centre of C is (11/2, -6) and r = √245/2 = 7√5/2. The diameter is 7√5, option (a).

The idea to keep. A circle touching line L at point A is always (point circle at A) + λ L. If a common chord must be a diameter of another circle, make it pass through that circle's centre.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 120 times in JEE Main since 2020. This question is from 2021.