# Thread: To find the equation of a sphere (in 3D space) given a plane, line and radius?

1. ## To find the equation of a sphere (in 3D space) given a plane, line and radius?

So given the plane: -x-2+2=5 and the line L: (2,1,0) + mu(-1,2,1) how can you find the 2 spheres that are tangent to the plane and whose centres pass through the line and who have radii 4?

2. Originally Posted by Yehia
So given the plane: -x-2+2=5 and the line L: (2,1,0) + mu(-1,2,1) how can you find the 2 spheres that are tangent to the plane and whose centres pass through the line and who have radii 4?
The equation of the plane looks odd. I'm guessing that it should actually be $-x-2y+2z = 5$.

The centre C of the sphere must lie on the line L, so $C = (2-\mu,1+2\mu,\mu)$ for some value of $\mu$. Also, the distance from C to the plane must be 4. The formula for the distance from the point $(x_0,y_0,z_0)$ to the plane $ax+by+cz=d$ is $\frac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}$. Plug the coordinates of C and the coefficients of the equation of the plane into that formula, set it equal to 4, and you have an equation for the two values of $\mu$. You then know the coordinates of C, and the rest is easy.