# Contact point of sphere and plane

This is a silly question but I'm stuck.. I'm given the sphere $(C):(x-1)^2+(y-2)^2+(z-3)^2=1$ with radius $r=1$ and center $K(1,2,3)$ and the plane $(P):x+y+z=6+\sqrt3$

I proved that they are tangent, I need to find their contact point

Any hints?

• you could use the general point $X$ on plane $P$ and then calculate the distance $\overline{XK}$. The point of contact then would be the minimum of that function. – Dr. Richard Klitzing May 12 '18 at 11:27

• Ok I thing I got it! $u=(1,1,1)$ is perpendicular to $P$ so the line that passes through $K$ perpendicular to $P$ is $x=t+1,y=t+2,z=t+3$ $(P): t+1+t+2+t+3=6+\sqrt3$ so $t=\sqrt3/3$ So $x=\sqrt3/3+1, y=\sqrt3/3+2, z=\sqrt3/3+3$ And the contact point is $(\sqrt3/3+1,\sqrt3/3+2,\sqrt3/3+3)$ Is this correct? – VakiPitsi May 12 '18 at 11:37
The contact point will be the closest point on the plane from the center of the sphere. So if the point is $(x,y,z)$ then we have to minimize the distance $(x-1)^2+(y-2)^2+(z-3)^2$ subject to the condition that the point belongs to the plane. Using Lagrange multiplier it is done quickly.