The equation of the circle passing through th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the ellipse equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$,we have $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.Eccentricity $e = \sqrt{1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the ellipse equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$,we have $a^2 = 16$ and $b^2 = 9$,so $a = 4$ and $b = 3$.Eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.The foci are at $(\pm ae, 0) = (\pm \sqrt{7}, 0)$.The circle passes through $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ and has its centre at $(0, 3)$.The radius $r$ of the circle is the distance from the centre $(0, 3)$ to a focus $(\sqrt{7}, 0)$:$r^2 = (\sqrt{7} - 0)^2 + (0 - 3)^2 = 7 + 9 = 16$,so $r = 4$.The equation of the circle with centre $(0, 3)$ and radius $r = 4$ is $(x - 0)^2 + (y - 3)^2 = 4^2$.$x^2 + y^2 - 6y + 9 = 16$.$x^2 + y^2 - 6y - 7 = 0$.

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

Similar Questions

Find the centre and radius of the circle $x^{2}+y^{2}-4x-8y-45=0$.

The equation of the circle that touches the $Y-$axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X-$axis is

If the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ are diameters of a circle with circumference $10\pi$,find the equation of the circle.

The points $(1,0), (0,1), (0,0)$ and $(2k, 3k), k \neq 0$ are concyclic,if $k$ is

Find the equation of the circle passing through $(1, -2)$ and touching the $x$-axis at $(3, 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module