# Give an equation which describes the intersection of this sphere with the plane?

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Lets say we have an equation of a sphere: (x+6)^2 + (y+4)^2 + (z-4)^2 = 16

Give an equation which describes the intersection of this sphere with the plane z=5

How would I go about doing this?

• 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 cross-sections.

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

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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 y-z plane), y = 0 (for the x-z plane), and z = 0 (for the x-y 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 x-z plane (x – 2)^2 + (y + 6)^2 = 25 – sixteen = 9

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