The radius of the circle in which the sphere x^2+y^2+z^2+2x-2y-4z-19=0 is cut by the plane…

Question

The radius of the circle in which the sphere $x^2+y^2+z^2+2x-2y-4z-19=0$ is cut by the plane $x+2y+2z+7=0$ is

Accepted Answer

Center of sphere $$ = \left( { - 1,1,2} \right)$$

Radius of sphere $$\sqrt {1 + 1 + 4 + 19}  = 5$$

Perpendicular distance from center to the plane

$$OC = d = \left| {{{ - 1 + 2 + 4 + 7} \over {\sqrt {1 + 4 + 4} }}} \right| = {{12} \over 3} = 4.$$

$$A{C^2} = A{O^2} - O{C^2} = {5^2} - {4^2} = 9$$

$$ \Rightarrow AC = 3$$

Solutions · 1