Question

Equation of the sphere which passes through the circle $x^2+y^2=4;z=0$ and is cut by the plane $x+2y+2z=0$ is a circle of radius $3$ is

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

Answer 1

Answer: (A)

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

The given circle is $x^2+y^2-4=0$,$z=0$

The equation of this circle can be written as $x^2+y^2+z^2-4=0$, $z=0$

Any sphere through this circle is $x^2+y^2+z^2-4+kz=0$ ...$\left(1\right)$

Its centre is $C$ $(0,0,\frac{-k}{2})$ and radius is $\sqrt{\frac{k^2}{4}+4}=$ $CP$

Now the sphere $\left(1\right)$ is cut by the plane $x+2y+2z=0$ ...$\left(2\right)$

In a circle of radius $3$, draw $CA\perp$ distance of $(0,0,\frac{-k}{2})$ from the plane (2) $=\frac{|0+0-k|}{\sqrt{1+4+4}}=\frac{k}{3}$

Now from the right $\angle d△CAP,{CA}^2+A{P}^2={CP}^2$

$\Rightarrow \frac{k^2}{9}+9=\frac{k^2}{4}+4$

$\Rightarrow k^2=36$

$\Rightarrow k=\pm 6$

Putting these values of $k$ in (1), the required sphere are $x^2+y^2+z^2\pm 6z-4=0$