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.