# equation of a plane touching a sphere

• Jun 11th 2009, 08:38 AM
Frostking
equation of a plane touching a sphere
I need to find the equation for a plane that just touches a sphere (x - 4)^2 + (y + 1)^2 + (z - 3)^2 = 4 that is parallel to the xy plane

In writing this I think I would just set x and y to zero and solve for z and therefore z = 5 would be the answer. Could someone please confirm this for me??

So, if this were to be parallel to the zy plane then I would just solve for x with z and y at 0, Right? IN that case x would = 2

THanks so much for all the assistance you folks have provided, it really helps. Frostking
• Jun 11th 2009, 09:37 AM
Soroban
Hello, Frostking!

This is easier than you think . . .

Quote:

Find the equation of a plane tangent to the sphere $\displaystyle (x - 4)^2 + (y + 1)^2 + (z - 3)^2 \:=\: 4$
that is parallel to the $\displaystyle xy$-plane

Try to visualize the graph . . .

We have a sphere: center (4, -1, 3) and radius 2.

A plane parallel to the $\displaystyle xy$-plane is "horizontal" .$\displaystyle (z \,=\,\text{constant}).$

It would be tangent to the sphere at its "north pole" or "south pole."

The north pole is (4, -1, 5); the south pole is (4,-1, 1).

Hence, the tangent plane is either: .$\displaystyle z = 5\:\text{ or }\:z = 1$