If you are given the sphere: x²+y²+z²+x-6z=0 and the plane x+y+=0
how do you find the planes which are parallell to that plane which intersect the sphere to create circles of radius 3.
Any help would be much appreciated. Thanks!
If you are given the sphere: x²+y²+z²+x-6z=0 and the plane x+y+=0
how do you find the planes which are parallell to that plane which intersect the sphere to create circles of radius 3.
Any help would be much appreciated. Thanks!
The circle is $\displaystyle (x+\frac12)^2 + y^2 + (z-3)^2 = \frac{37}4$, with centre at $\displaystyle (-\frac12,0,3)$ and radius $\displaystyle \sqrt{\frac{37}4}$. A plane that intersects it in a circle of radius 3 must have distance $\displaystyle \sqrt{\frac{37}4 - 3^2} = \frac12$ from the centre of the sphere (by Pythagoras). So you are looking for a plane with equation $\displaystyle x+y+z=k$, where k is chosen so that the distance from $\displaystyle (-\frac12,0,3)$ to the plane is 1/2.