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JEE Main 2013

JEE Main 2013 Mathematics question on Conic Sections

The equation of the circle passing through the foci of the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 1 and having centre at (0,3)(0, 3) is
  1. A.
    x2+y2−6y−7=0x^2 + y^2 - 6y - 7 = 0
  2. B.
    x2+y2−6y+7=0x^2 + y^2 - 6y + 7 = 0
  3. C.
    x2+y2−6y−5=0x^2 + y^2 - 6y - 5 = 0
  4. D.
    x2+y2−6y+5=0x^2 + y^2 - 6y + 5 = 0
Show answer and solution

Correct option: A

Solution

For the ellipse x²/16 + y²/9 = 1, a²=16, b²=9, so c²=16-9=7, giving foci at (±√7, 0).

The radius of the circle centered at (0,3) passing through (√7, 0) is r = √((√7-0)² + (0-3)²) = √(7+9) = √16 = 4.

Circle equation: x² + (y-3)² = 16 → x² + y² - 6y + 9 - 16 = 0 → x² + y² - 6y - 7 = 0.

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