Question

Find the equation of the sphere with center $(3,6,-4)$ and touching the plane $2x-2y-z-10=0$?

Answer 1

Step 1: Find the radius of the sphere

Let $r$ is the radius of sphere

The radius of the sphere = Distance between the center of the sphere and the point at which the plane touches the sphere

Distance between a point$\left(x_1,y_1,z_1\right)$ and a plane$(ax+by+cz+d=0)$ is given as

$r=\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}$

$\Rightarrow$ $=\left|\frac{2\times 3-2\times 6+4-10}{\sqrt{{\left(2\right)}^2+{\left(-2\right)}^2+1}}\right|$

$\Rightarrow$ $=\frac{12}{\sqrt{9}}$

$\Rightarrow$ $=\frac{12}{3}$

$\Rightarrow$ $r=4$

Step 2: Solve for the equation of sphere using center radius form of equation

The equation of sphere is

${\left(x-a\right)}^2+{\left(y-b\right)}^2+{\left(z-c\right)}^2=r^2$ where $\left(a,b,c\right)$ are the co-ordinates of the center of the sphere

$\Rightarrow$ ${\left(x-3\right)}^2+{\left(y-6\right)}^2+{\left(z+4\right)}^2=4^2$

$\Rightarrow x^2+y^2+z^2-6x-12y+8z+45=0$

Hence, the required equation of the sphere whose center is $(3,6,-4)$ and $r=4$is $x^2+y^2+z^2-6x-12y+8z+45=0$