jee-main 2013 Q73

jee-main · India · 07apr Circles Circle Equation Derivation

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
(1) $x^2 + y^2 - 6y - 5 = 0$
(2) $x^2 + y^2 - 6y + 5 = 0$
(3) $x^2 + y^2 - 6y - 7 = 0$
(4) $x^2 + y^2 - 6y + 7 = 0$

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\\
(1) $x^2 + y^2 - 6y - 5 = 0$\\
(2) $x^2 + y^2 - 6y + 5 = 0$\\
(3) $x^2 + y^2 - 6y - 7 = 0$\\
(4) $x^2 + y^2 - 6y + 7 = 0$

Paper Questions

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