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!

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- Sep 26th 2010, 05:19 AMYehiaFinding planes that intersect spheres in 3D vector Geometry
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! - Sep 26th 2010, 09:37 AMOpalg
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.