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) is: - Mathematics MCQ JEE Main 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

1. Ellipse: \frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1 . Here a^{2} = 16, b^{2} = 9 .2. Foci distance from center: c^{2} = a^{2} - b^{2} = 16 - 9 = 7 .3. Foci are (\pm 7, 0) .4. Circle center is (0, 3) . Radius R^{2} = (7 - 0)^{2} + (0 - 3)^{2} = 7 + 9 = 16 .5. Circle equation: (x - 0)^{2} + (y - 3)^{2} = 16 . x^{2} + y^{2} - 6 y + 9 = 16 x^{2} + y^{2} - 6 y - 7 = 0

Try it yourself before checking the explanation above.

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

x^2 + y^2 - 6y + 7 = 0

x^2 + y^2 - 6y - 5 = 0

x^2 + y^2 - 6y + 5 = 0

x^2 + y^2 - 6y - 7 = 0

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