A sphere consists of all the points that are 5 units from its center

(2, 3,−6). Write an equation that describes this sphere. Does the sphere intersect the xy-plane? Explain.

Why did the sphere bring a map to the party? Because it didn't want to get lost in space!

To write the equation for the sphere, we can use the formula:

(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2

Where (a, b, c) represents the center of the sphere and r represents the radius.

In this case, the center of the sphere is (2, 3, -6) and the radius is 5 units. Substituting these values into the equation, we get:

(x - 2)^2 + (y - 3)^2 + (z + 6)^2 = 5^2

Simplifying this equation, we have:

(x - 2)^2 + (y - 3)^2 + (z + 6)^2 = 25

Now, to determine if the sphere intersects the xy-plane, we can substitute z = 0 into the equation and see if we get any valid solutions for x and y.

Substituting z = 0, we have:

(x - 2)^2 + (y - 3)^2 + (0 + 6)^2 = 25

Simplifying further, we get:

(x - 2)^2 + (y - 3)^2 + 6^2 = 25

(x - 2)^2 + (y - 3)^2 = 25 - 36

(x - 2)^2 + (y - 3)^2 = -11

Uh oh! We have a bit of a problem here. The equation (x - 2)^2 + (y - 3)^2 = -11 doesn't have any valid solutions, which means the sphere doesn't intersect the xy-plane. So, the sphere is floating happily above the xy-plane, like a balloon at a birthday party!

To write the equation of the sphere, we can use the formula:

(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2

Where (h, k, l) represents the center of the sphere and r represents the radius.

From the given information, we know that the center of the sphere is (2, 3, -6) and the radius is 5 units. Substituting these values into the formula, we get:

(x - 2)^2 + (y - 3)^2 + (z - (-6))^2 = 5^2

Simplifying further, we have:

(x - 2)^2 + (y - 3)^2 + (z + 6)^2 = 25

Therefore, the equation that describes this sphere is (x - 2)^2 + (y - 3)^2 + (z + 6)^2 = 25.

To determine whether the sphere intersects the xy-plane, we need to check if there are any z-values in the equation that would result in z = 0 (the xy-plane).

Substituting z = 0 into the equation, we get:

(x - 2)^2 + (y - 3)^2 + (0 + 6)^2 = 25

(x - 2)^2 + (y - 3)^2 + 6^2 = 25

(x - 2)^2 + (y - 3)^2 + 36 = 25

(x - 2)^2 + (y - 3)^2 = 25 - 36

(x - 2)^2 + (y - 3)^2 = -11

Since the sum of squares cannot be negative, there are no points that satisfy this equation. Therefore, the sphere does not intersect the xy-plane.

To write the equation of the sphere, we need to use the general equation of a sphere:

(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2

Where (h, k, l) represents the coordinates of the center of the sphere, and r represents its radius.

In this case, the given coordinates of the center of the sphere are (2, 3, -6), and the radius is 5 units.

Plugging in these values, we get:

(x - 2)^2 + (y - 3)^2 + (z - (-6))^2 = 5^2

Simplifying further, we have:

(x - 2)^2 + (y - 3)^2 + (z + 6)^2 = 25

This is the equation that describes the sphere.

To determine whether the sphere intersects the xy-plane, we can examine the equation. The xy-plane is a plane where z = 0.

Plugging in z = 0 into the equation, we get:

(x - 2)^2 + (y - 3)^2 + (0 + 6)^2 = 25

(x - 2)^2 + (y - 3)^2 + 36 = 25

(x - 2)^2 + (y - 3)^2 = -11

As you can see, this equation does not have a real solution because the left-hand side is the sum of two squares and because of this, must be greater than or equal to zero.

Therefore, the sphere does not intersect the xy-plane because the equation does not have any real solutions on the xy-plane (z=0).

(x-2)^2 + (y-3)^2 + (z+6)^2 = 25

is the equation. That is the general equation for a sphere with the center coordinates and radius prescribed.

That should be obvious.

If it intersected the xy plane, there would be a solution for z = 0. That would require that
(x-2)^2 + (y-3)^2 + 36 = 25
which is impossible for real x and y.

You would be have a sphere of radius 5 with its center 6 units below the plane.