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AIEEE 2011

JEE Main 2011 Mathematics question on Circles

The circle passing through the point (−1,0)(-1, 0) and touching the Y-axis at (0,2)(0, 2) also passes through the point
  1. A.
    (−32,0)\left(-\dfrac{3}{2}, 0\right)
  2. B.
    (−52,2)\left(-\dfrac{5}{2}, 2\right)
  3. C.
    (−32,52)\left(-\dfrac{3}{2}, \dfrac{5}{2}\right)
  4. D.
    (−4,0)(-4, 0)
Show answer and solution

Correct option: D

Solution

The circle touches the Y-axis at (0,2), so its center must be at (a, 2) for some a, and its radius = |a|. The equation is (x−a)² + (y−2)² = a².

Since it passes through (−1, 0): (−1−a)² + 4 = a², giving 1 + 2a + a² + 4 = a², so 2a = −5, a = −5/2.

Center is (−5/2, 2), radius = 5/2. Check (−4, 0): (−4+5/2)² + (0−2)² = (−3/2)² + 4 = 9/4 + 16/4 = 25/4 = (5/2)². ✓

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