Find the equation of the sphere which passes through the circle $x^{2} + y^{2} = 4$ , z = 0 and is cut by the plane x+2y +2z =0 in a circle of radius 3.

Open in App

Solution

general equation of sphere,
$x^{2} + y^{2} + 2 g x + 2 f y + 2 h z + c = 0$
centre $(- g, - f . - h)$ and radius $= \sqrt{g^{2} + f^{2} + h^{2} - c}$
At $z = 0$ , $x^{2} + y^{2} + 2 g x + 2 f y + c = 0 - - - (i)$
$x^{2} + y^{2} - 4 = 0 - - - (i i)$
comparing $(i)$ and $(i i)$ , $g = 0, f = 0, c = - 4$
$∴$ general equation reduces to,
$x^{2} + y^{2} + z^{2} + 2 h z + 4 = 0$ centre $(0, 0, h)$ , radius $= \sqrt{h^{2} - c}$
from fig., $P$ is perpendicular drawn from centre to plane and $r$ is radius of sphere. $∴ p^{2} + 9 = r^{2}$
$\frac{4 h^{2}}{9} + 9 = h^{2} + 4$
$\Rightarrow h = \pm 3$
$∴$ equation of spheres are,
$x^{2} + y^{2} + z^{2} + 6 z - 4 = 0$ and $x^{2} + y^{2} + z^{2} + 6 z - 4 = 0$

Suggest Corrections

Similar questions

x^{2} + y^{2} = 4; z = 0

x + 2 y + 2 z = 0

3

x^{2} + y^{2} = 4, z = 0

x + 2 y + 2 z = 0

3

x^{2} + y^{2} + z^{2} + 2 z - 2 y - 4 z - 19 = 0

x + 2 y + 2 z + 7 = 0

x + 2 y - z = 4

x^{2} + y^{2} + z^{2} - x + z - 2 = 0

x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0

Join BYJU'S Learning Program