Lets say we have an equation of a sphere: (x+6)^2 + (y+4)^2 + (z4)^2 = 16
Give an equation which describes the intersection of this sphere with the plane z=5
How would I go about doing this?
3 Answers

By plugging in the equation of the plane (z = 5) into the equation of the sphere, we see that the equation of intersection is the following circle at height z = 5:
(x + 6)^2 + (y + 4)^2 + (5 – 4)^2 = 16 ==> (x + 6)^2 + (y + 4)^2 = 15;
this should make sense since any plane that intersects a sphere parallel to a diameter results in circular crosssections.
I hope this helps!

Equation of Sphere:
(x + 6)² + (y + 4)² + (z – 4)² = 16
Equation of Plane:
z = 5
Subbing 5 for z in the Equation of Sphere,
Equation of Intersecting Circle:
(x + 6)² + (y + 4)² + (5 – 4)² = 16
(x + 6)² + (y + 4)² + (1)² = 16
(x + 6)² + (y + 4)² + 1 = 16
(x + 6)² + (y + 4)² = 16 – 1
(x + 6)² + (y + 4)² = 15
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Source(s): 9/6/12 
the middle of the sector are: (2,6,4) subsequently, the equation is given via: (x – 2)^2 + (y + 6)^2 + (z – 4)^2 = 25 The intersections are stumbled on whilst x = 0 (for the yz plane), y = 0 (for the xz plane), and z = 0 (for the xy plane). Now examine: (y + 6)^2 + (z – 4)^2 = 25 – 4 = 21 (x – 2)^2 + (z – 4)^2 = 25 – 36 < 0 so no intersection with the xz plane (x – 2)^2 + (y + 6)^2 = 25 – sixteen = 9