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find an equation of sphere centered at (-10,-7,1) with radius 3 give an equation which describes the intersection of this sphere with plane z=2

User Bendystraw
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1 Answer

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Final answer:

The sphere centered at (-10, -7, 1) with radius 3 has the equation (x + 10)^2 + (y + 7)^2 + (z - 1)^2 = 9.

The circle representing the intersection with the plane z = 2 has the equation (x + 10)^2 + (y + 7)^2 = 8, and its center is (-10, -7, 2) with radius √8.

Step-by-step explanation:

The equation of a sphere centered at a point (h, k, l) with radius r is (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2.

In this case, the center is (-10, -7, 1) and the radius is 3, so the equation of the sphere is as follows:

(x + 10)^2 + (y + 7)^2 + (z - 1)^2 = 9.

To find the equation of the intersection of this sphere with the plane z = 2, we substitute z with 2 in the equation of the sphere, giving:

(x + 10)^2 + (y + 7)^2 + (2 - 1)^2 = 9.

Simplify this equation to get:

(x + 10)^2 + (y + 7)^2 = 8.

This represents a circle on the plane z = 2 with center (-10, -7, 2) and radius √8.

User John Wales
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