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JEE Main 20137 April
Exam
JEE Main
Year
2013
Date
7 April
Subject
Mathematics
Chapter
Ellipse

JEE Main 2013 · 7 April · Mathematics · Ellipse

The equation of the circle passing through the foci of the ellipse...

The equation of the circle passing through the foci of the ellipse x216+y29=1\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{9} = 1, and having centre at (0,3)\left(0, 3\right) isThe equation of the circle passing through the foci of the ellipse x ^ 2 16 + y ^ 2 9 = 1 , and having centre at (0, 3)

Options

  1. Ax2+y2−6y−5=0{x}^{2} + {y}^{2} - 6y - 5 = 0
  2. Bx2+y2−6y+5=0{x}^{2} + {y}^{2} - 6y + 5 = 0
  3. Cx2+y2−6y−7=0{x}^{2} + {y}^{2} - 6y - 7 = 0
  4. Dx2+y2−6y+7=0{x}^{2} + {y}^{2} - 6y + 7 = 0

Correct answer

C. x2+y2−6y−7=0{x}^{2} + {y}^{2} - 6y - 7 = 0

Step-by-step solution

For x216+y29=1\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{9} = 1

The equation of the circle passing through the foci of the ellipse x ^ 2 16 + y ^ 2 9 = 1 , and having centre at (0, 3)

ae=16−9=7ae = \sqrt{16 - 9} = \sqrt{7}

Foci=(±7,0)= \left(\pm \sqrt{7}, 0\right)

Radius=7+9=4= \sqrt{7 + 9} = 4

∴∴Equation of circle=(x−0)2+(y−3)2=42= {\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2} = {4}^{2}

⇒x2+y2−6y−7=0⇒ {x}^{2} + {y}^{2} - 6y - 7 = 0

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