Find an equation of the sphere with center (2,-6,4) and radius 5. Describe it intersection with each of the coordinate planes.

you know that a sphere with center at (0,0,0) is x^2 + y^2 + z^2 = 25

So, shift the center and you have
(x-2)^2 + (y+6)^2 + (z-4)^2 = 25
To find the intersection with the xy plane, set z=0 and you have
(x-2)^2 + (y+6)^2 + (0-4)^2 = 25
(x-2)^2 + (y+6)^2 + 16 = 25
(x-2)^2 + (y+6)^2 = 9
So, it's a circle of radius 3 with center at (2,-6)

and so on for the other planes.

Well, well, well, if it isn't the sphere with all its fancy center and radius! Let's have some fun and find an equation for this little guy.

The equation of a sphere with center (a, b, c) and radius r is (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2. So, plugging in the values from our center (2, -6, 4) and radius 5, we get:

(x - 2)^2 + (y + 6)^2 + (z - 4)^2 = 25.

Now let's see what happens when this sphere meets the coordinate planes!

When a sphere intersects the xy-plane, the z-coordinate is always 0. So, we can set z = 0 in the equation of the sphere and simplify it:

(x - 2)^2 + (y + 6)^2 + (-4)^2 = 25,
(x - 2)^2 + (y + 6)^2 + 16 = 25,
(x - 2)^2 + (y + 6)^2 = 9.

This equation represents a circle on the xy-plane with center (2, -6) and radius 3.

When it comes to the intersection with the xz-plane, the y-coordinate should be 0. So, we set y = 0 in the sphere's equation:

(x - 2)^2 + (0 + 6)^2 + (z - 4)^2 = 25,
(x - 2)^2 + 36 + (z - 4)^2 = 25,
(x - 2)^2 + (z - 4)^2 = -11.

Uh-oh! We've got a bit of a problem here. The correct answer is that there is no intersection between the sphere and the xz-plane. I guess the sphere just isn't a fan of the xz-plane.

Lastly, we'll take a look at the intersection with the yz-plane. This time, we let x = 0 in the equation:

(0 - 2)^2 + (y + 6)^2 + (z - 4)^2 = 25,
4 + (y + 6)^2 + (z - 4)^2 = 25,
(y + 6)^2 + (z - 4)^2 = 21.

This equation represents another circle on the yz-plane, center (-6, 4) with radius √21.

So, there you have it! The sphere cuts a circle on the xy-plane with radius 3, and a circle on the yz-plane with radius √21, but it turns its back on the xz-plane. I guess not every plane is good enough for a sphere's attention.

To find the equation of a sphere with center (h, k, l) and radius r, the equation is:

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

In this case, the center is (2, -6, 4) and the radius is 5. So, the equation of the sphere is:

(x - 2)^2 + (y + 6)^2 + (z - 4)^2 = 25

Now, let's examine the intersection of the sphere with each of the coordinate planes:

1. Intersection with the xy-plane (z = 0):
To find this intersection, we substitute z = 0 into the equation of the sphere:
(x - 2)^2 + (y + 6)^2 + (0 - 4)^2 = 25
Simplifying this equation yields:
(x - 2)^2 + (y + 6)^2 + 16 = 25
(x - 2)^2 + (y + 6)^2 = 9

Therefore, the intersection with the xy-plane is the circle with center (2, -6) and radius 3.

2. Intersection with the yz-plane (x = 0):
To find this intersection, we substitute x = 0 into the equation of the sphere:
(0 - 2)^2 + (y + 6)^2 + (z - 4)^2 = 25
Simplifying this equation yields:
4 + (y + 6)^2 + (z - 4)^2 = 25
(y + 6)^2 + (z - 4)^2 = 21

Therefore, the intersection with the yz-plane is an ellipse centered at (0, -6, 4) with major axis along the y-axis and minor axis along the z-axis.

3. Intersection with the xz-plane (y = 0):
To find this intersection, we substitute y = 0 into the equation of the sphere:
(x - 2)^2 + (0 + 6)^2 + (z - 4)^2 = 25
Simplifying this equation yields:
(x - 2)^2 + 36 + (z - 4)^2 = 25
(x - 2)^2 + (z - 4)^2 = -11

Since the equation has no solution, the intersection with the xz-plane is empty.

In summary:
- The intersection with the xy-plane is a circle with center (2, -6) and radius 3.
- The intersection with the yz-plane is an ellipse centered at (0, -6, 4) with major axis along the y-axis and minor axis along the z-axis.
- The intersection with the xz-plane is empty.

To find an equation of the sphere with a given center and radius, we can use the formula:

(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 the radius of the sphere. In this case, the center is (2,-6,4) and the radius is 5, so the equation of the sphere is:

(x - 2)^2 + (y + 6)^2 + (z - 4)^2 = 5^2

Simplifying this equation gives:

(x - 2)^2 + (y + 6)^2 + (z - 4)^2 = 25

Now, let's describe the intersection of the sphere with each of the coordinate planes:

1. Intersection with the xy-plane (z = 0):
Substitute z = 0 into the equation of the sphere:
(x - 2)^2 + (y + 6)^2 + (0 - 4)^2 = 25
(x - 2)^2 + (y + 6)^2 + 16 = 25
(x - 2)^2 + (y + 6)^2 = 9

So, the intersection of the sphere with the xy-plane is a circle with center (2, -6) and radius 3.

2. Intersection with the yz-plane (x = 0):
Substitute x = 0 into the equation of the sphere:
(0 - 2)^2 + (y + 6)^2 + (z - 4)^2 = 25
4 + (y + 6)^2 + (z - 4)^2 = 25
(y + 6)^2 + (z - 4)^2 = 21

So, the intersection of the sphere with the yz-plane is an ellipse centered at (-6, 4) with major axis along the y-axis.

3. Intersection with the xz-plane (y = 0):
Substitute y = 0 into the equation of the sphere:
(x - 2)^2 + (0 + 6)^2 + (z - 4)^2 = 25
(x - 2)^2 + 36 + (z - 4)^2 = 25
(x - 2)^2 + (z - 4)^2 = -11

Since the sum of two squares cannot be negative, there is no intersection of the sphere with the xz-plane.

Therefore, the intersection of the sphere with the xy-plane is a circle, the intersection with the yz-plane is an ellipse, and there is no intersection with the xz-plane.