Determine the equation of the sphere that is tangent to the plane
$$(P):6x-3y-2z-35=0$$ at $M(5,-1,-1)$ and also tangent to the plane
$$(Q):6x-3y-2z+63=0$$
My solution:
Let $d$ be the line passing through $M$, $d$ orthogonal to $(P)$. The parametric equations of the line $d$ are:
$$x = 5+6t$$
$$y=-1-3t$$
$$z=-1-2t$$
So the center of the sphere is of the form $C(5+6t, -1-3t,-1-2t)$. The center of the sphere must belong to the line $d$ and the plane $(Q)$, so it is the intersection point between the line and the plane.
$$6(5+6t)+3(1+3t)+2(1+2t)+63=0$$
From this relation we get $t=2$, so the center has the coordinates $C(17,-7,-5)$. Now since it's stated that the plane $(Q)$ is tangent to the sphere, the distance from the center to the plane is equal with the radius. The radius $r=14$, so now we have the equation of the sphere:
$$(x-17)^2+(y+7)^2+(z+5)^2=14^2$$
Is this correct? I'm not sure if my initial condition for the center to belong to the line $d$ and the plane $(Q)$ is right, but I based my assumption off of the fact that they are both tangent to the sphere and so the distance from the center to each plane is equal with the radius.
EDIT: Okay, so my assumption was definitely wrong, I somehow didn't notice the normal vectors were identical
Best Answer
No, that is not correct. You already know that the center of the sphere is of the form $(5+6t,-1-3t,-1-2t)$, for some real $t$. Since the planes $P$ and $Q$ and parallel, $C$ has to be the midpoint between $M$ and the point $M'$ which belongs to the line $(5+6t,-1-3t,-1-2t)$ and to $Q$. But $M'=(-7,5,3)$, and therefore $C=(-1,2,1)$ (and the radius is $7$).