There is a plane given the equation. If a sphere is tangent to this plane, what is the equation for the sphere?(Sphere Center: ${M(4,-2.3)}$)

Question

This is plane equation:
${-2x+y-2z+7=0}$

What is the equation for the sphere tangent to this plane? Center of the sphere = ${M(4,-2,3)}$

Sphere equation form:
${(x-a)^{2}+(y-b)^{2}+(z-c)^{2}=r^{2}}$

If center values are placed in place:
${(x-4)^{2}+(y+2)^{2}+(z-3)^{2}=r^{2}}$

So, Since the sphere is tangent to the plane, they intersect at a common point, and this intersection ${P(x,y,z)}$ point must satisfy both the plane equation and the sphere equation, right?
I request your help.

Answer 1

The norm of the normal vector $(-2, 1, 2)$ is $3$. Hence the radius of the sphere is \begin{equation} r = |-2(4)+(-2)-2(3)+7|/3 = 9/3=3 \end{equation} The point where they cross is $(4, -2, 3) \pm (-2, 1, -2)r/3 = (2, -1, 1)$.