Question

Equation of the sphere with center in the positive octant 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 $3$ is

• A. $x^2+y^2+z^2-6x-4=0$
• B. $x^2+y^2+z^2-6z+4=0$
• C. $x^2+y^2+z^2-6z-4=0$
• D. $x^2+y^2+z^2-6y-4=0$

Answer 1

The correct option is C $x^2+y^2+z^2-6z-4=0$

Equation of a sphere through the given circle is

$x^2+y^2+z^2-4\lambda +z=0$

Centre of the sphere is $(0,0-\frac{\lambda }{2})$ and the radius

$r=\sqrt{0+0+\frac{{\lambda }^2}{4}+4}$

Let $d$ the distance of the plane $x+2y+2z=0$ from the centre of the sphere.

then, $d=\left|\frac{0+2\times 0+2(-\frac{\lambda }{2})}{\sqrt{1+2^2+2^2}}\right|=\left|\frac{\lambda }{3}\right|.$

Since the sphere (1) cuts the plane in a circle of radius $3,$

$r^2-d^2=3^2\Rightarrow \frac{{\lambda }^2}{4}+4-\frac{{\lambda }^2}{9}=9\Rightarrow \frac{5{\lambda }^2}{36}=5$

$\Rightarrow {\lambda }^2=36\Rightarrow \lambda =\pm 6$

Since the centre lies in the positive octant, $\lambda <0$ and hence the required equation is

$x^2+y^2+z^2-6z-4=0$