The equation of the circle passing through the foci of the ellipse, and having centre at $(0, 3)$ is

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x^{2} + y^{2} - 6 y - 7 = 0$
The equation of the circle passing through the foci of the ellipse and having centre as $(0, 3)$ is
$x^{2} + y^{2} - 6 y - 7 = 0$

Suggest Corrections

Similar questions

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

(0, 3)

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

(0, 3)

x^{2} + y^{2} - 8 x + 6 y - 5 = 0

(- 2, - 7)

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

x^{2} + y^{2} - 4 x - 6 y = 0

The diameters of a circle are along $2 x + y - 7 = 0$ and $x + 3 y - 11 = 0$ Then the equation of this circle, which also passes through (5,7) is

Join BYJU'S Learning Program