Find the equation of the sphere given the points P (2, 3, 4) and Q(−4, 1, 2) are endpoints of a diameter of the sphere. In what way does this sphere intersect the yz-plane, or explain why it does not intersect the yz-plane.

### Finding the Equation of the Sphere

To find the equation of the sphere, we use the following formula for the equation of a sphere:

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

Where:
- \((h, k, l)\) is the center of the sphere,
- \(r\) is the radius of the sphere.

#### Step 1: Find the center of the sphere
The points \(P(2, 3, 4)\) and \(Q(-4, 1, 2)\) are the endpoints of a diameter of the sphere. The center of the sphere is the midpoint of the diameter, which can be calculated as:

\[
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)
\]

Substitute the coordinates of \(P(2, 3, 4)\) and \(Q(-4, 1, 2)\):

\[
\left( \frac{2 + (-4)}{2}, \frac{3 + 1}{2}, \frac{4 + 2}{2} \right) = \left( \frac{-2}{2}, \frac{4}{2}, \frac{6}{2} \right) = (-1, 2, 3)
\]

So, the center of the sphere is \((-1, 2, 3)\).

#### Step 2: Find the radius of the sphere
The radius of the sphere is half the length of the diameter. The length of the diameter is the distance between points \(P(2, 3, 4)\) and \(Q(-4, 1, 2)\). The formula for the distance between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}
\]

Substitute the coordinates of \(P(2, 3, 4)\) and \(Q(-4, 1, 2)\):

\[
d = \sqrt{(-4 - 2)^2 + (1 - 3)^2 + (2 - 4)^2} = \sqrt{(-6)^2 + (-2)^2 + (-2)^2} = \sqrt{36 + 4 + 4} = \sqrt{44}
\]

The length of the diameter is \(\sqrt{44}\), so the radius is:

\[
r = \frac{\sqrt{44}}{2} = \frac{2\sqrt{11}}{2} = \sqrt{11}
\]

#### Step 3: Equation of the sphere
The equation of the sphere is:

\[
(x - (-1))^2 + (y - 2)^2 + (z - 3)^2 = (\sqrt{11})^2
\]

Simplifying:

\[
(x + 1)^2 + (y - 2)^2 + (z - 3)^2 = 11
\]

Thus, the equation of the sphere is:

\[
(x + 1)^2 + (y - 2)^2 + (z - 3)^2 = 11
\]

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### Intersection with the \(yz\)-plane

The \(yz\)-plane is defined by \(x = 0\). To check how the sphere intersects the \(yz\)-plane, we substitute \(x = 0\) into the equation of the sphere:

\[
(0 + 1)^2 + (y - 2)^2 + (z - 3)^2 = 11
\]

This simplifies to:

\[
1 + (y - 2)^2 + (z - 3)^2 = 11
\]

\[
(y - 2)^2 + (z - 3)^2 = 10
\]

This is the equation of a circle in the \(yz\)-plane with center \((2, 3)\) and radius \(\sqrt{10}\). Hence, the sphere intersects the \(yz\)-plane in a circle.

---

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. How would you find the equation of a sphere if only one point on the sphere and the center were given?
2. Can a sphere have an empty intersection with the \(yz\)-plane? Under what conditions?
3. What happens to the equation of the sphere if the radius doubles?
4. How would the equation of the sphere change if it were translated along the \(x\)-axis by 5 units?
5. What is the geometric interpretation of the intersection of a sphere with any plane?

**Tip**: When analyzing how a sphere intersects a plane, the result will either be a circle, a point, or no intersection, depending on the distance between the center of the sphere and the plane.

Learn how to find the equation of a sphere using endpoints of its diameter at points P (2, 3, 4) and Q(−4, 1, 2). Discover how the sphere intersects the yz-plane, with a detailed step-by-step solution involving the midpoint and distance formulas. Suitable for high school students in Grades 10-12.

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