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JEE Main 2019 · 10 April · Shift I

JEE Main 2019 Mathematics question on Circles

The line x=yx = y touches a circle at the point (1,1)(1, 1). If the circle also passes through the point (1,−3)(1, -3), then its radius is
  1. A.
    323\sqrt{2}
  2. B.
    222\sqrt{2}
  3. C.
    22
  4. D.
    33
Show answer and solution

Correct option: B

Solution

Since x = y is tangent at (1,1), the radius at that point is perpendicular to x = y, so the center lies on the line y - 1 = -1(x - 1), i.e., y = -x + 2. Let center = (a, 2-a).

Equal distances from center to (1,1) and (1,-3):

(a-1)² + (1-a)² = (a-1)² + (5-a)²

Wait, let me redo: distance² to (1,1) = distance² to (1,-3):

(a-1)² + (2-a-1)² = (a-1)² + (2-a+3)²

(a-1)² + (1-a)² = (a-1)² + (5-a)²

(1-a)² = (5-a)²

1 - 2a + a² = 25 - 10a + a²

8a = 24 → a = 3, so center = (3, -1).

Radius = distance from (3,-1) to (1,1) = √(4+4) = 2√2.

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