Originally Posted by

**Schdero** I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time(Worried)

The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:

Sphere S: x^2+y^2+z^2-2y-22z-103=0

Point C, centre of the circle of intersection: 6/7/8

and the plane P 2x+2y-z-18=0

My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)

Then I invented the line l, which goes from C farther on the plane an eventually hits the spehere.

l= (6,7,8)+t(1,-1,0)

To find where it hits the plane, I inserted the line in the equation of the sphere

$\displaystyle ({\color{red}1}+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0$

As result, I get t^2=(70)^1/2