find the equation of a sphere with endpoints A and B where B is the point of tangency of the sphere and the plane

Question

Find the equation of a sphere with a diameter that has endpoints $A(1, 8, −2)$ and $B$, where $B$ is the point of tangency of the sphere with the plane $−9x +6y + 2z = 2$.

Now i know that i can get the distance of the point to the plane by $\frac{|-9(1)+6(8)+2(2)-2|}{\sqrt{81+36+4}}$ then dividing it by $2$ to get the radius. my problem then is finding the center of the sphere. can anybody help me???

Answer 1

Sure. The direction perpendicular to a plane with an equation like $$ ax + by + cz = d $$ is always $(a, b, c)$. That means that in your case, this direction is $(-9, 6, 2)$. So the ray from $A$ towards that plane consists of point of the form $$ (x, y, z) = (1, 8, -2) + t (-9, 6, 2) = (1 - 9t, 8-6t, -2+2t) $$ for any positive value of $t$.

Exactly ONE of these will lie on the plane, i.e., if you plug $$ (x, y, z) = (1 - 9t, 8-6t, -2+2t) $$ into the plane equation $$ -9x + 6y + 2z = 2 $$ and solve for $t$, you'll get just one answer. Suppose it's $t = 5$. Then your point $B$ is just $$ (1 - 9\cdot 5, 8-6\cdot 5, -2+2 \cdot 5). $$ And the sphere's center is just the average of $A$ and $B$.

Now...can you work out those steps for yourself?

Answer 2

Shouldn't it be $(1,8,-2)-1.5\cdot(-9,6,2)/11$, since the radius equals $3$ and the plane's normal vector points towards $A$?