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.

### Step 1: Find the center of the sphere
The center of the sphere lies at the midpoint of the diameter. Since \( P(2, 3, 4) \) and \( Q(-4, 1, 2) \) are endpoints of the diameter, the midpoint \( C \) can be calculated as:

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

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

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

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

### Step 2: Calculate the radius of the sphere
The radius is half the length of the diameter, so we first calculate the distance between points \( P \) and \( Q \) using the distance formula:

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

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

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

Thus, the diameter is \( \sqrt{44} \), and the radius is:

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

### Step 3: Write the equation of the sphere
The general equation of a sphere with center \( C(x_0, y_0, z_0) \) and radius \( r \) is:

\[
(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2
\]

Substitute \( C(-1, 2, 3) \) and \( r = \sqrt{11} \):

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

Thus, the equation of the sphere is:

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

### Step 4: Determine if the sphere intersects the \( yz \)-plane
The \( yz \)-plane is defined by \( x = 0 \). To see how the sphere intersects the \( yz \)-plane, substitute \( x = 0 \) into the equation of the sphere:

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

Simplifying:

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

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

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

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

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### Summary:
- The equation of the sphere is \( (x + 1)^2 + (y - 2)^2 + (z - 3)^2 = 11 \).
- The sphere intersects the \( yz \)-plane in a circle with center \( (2, 3) \) and radius \( \sqrt{10} \).

Do you want more details or have any questions?

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**Related Questions:**
1. How do you calculate the midpoint of a segment in 3D geometry?
2. What is the general formula for the distance between two points in 3D space?
3. How can you derive the equation of a sphere if only given its center and radius?
4. What is the geometric interpretation of a sphere intersecting a plane?
5. How can you find the equation of a circle formed by the intersection of a sphere and a plane?

**Tip:** Always verify the intersection of a surface with a plane by substituting the appropriate coordinates of the plane into the surface equation.

This problem explores how to find the equation of a sphere given endpoints P(2, 3, 4) and Q(−4, 1, 2) of a diameter, using midpoint and distance formulas. The problem also examines how the sphere intersects the yz-plane, showing the formation of a circle with a specific radius. Suitable for high school students in Grades 10-12.

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