Question
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Find the equation of the sphere passing through \( P(-2,3,4) \) and \( \mathrm{Q}(6,-1,3) \) with its center at the midpoint of PQ . The standard equation of the sphere is (Simplify your answer.)

Ask by Flynn Warner. in the United States
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The equation of the sphere is \( (x - 2)^2 + (y - 1)^2 + (z - 3.5)^2 = 20.25 \).

Solution

To find the equation of the sphere that passes through the points \( P(-2,3,4) \) and \( Q(6,-1,3) \) with its center at the midpoint of \( P \) and \( Q \), we can follow these steps: 1. **Calculate the midpoint of \( P \) and \( Q \)**: The midpoint \( M \) of two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) is given by: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \] 2. **Calculate the radius of the sphere**: The radius \( r \) of the sphere is the distance from the center \( M \) to either point \( P \) or \( Q \). The distance \( d \) between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] 3. **Write the standard equation of the sphere**: The standard equation of a sphere with center \( (h, k, l) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] Now, let's perform the calculations step by step. ### Step 1: Calculate the midpoint \( M \) Given: - \( P(-2, 3, 4) \) - \( Q(6, -1, 3) \) Calculating the midpoint: \[ M = \left( \frac{-2 + 6}{2}, \frac{3 - 1}{2}, \frac{4 + 3}{2} \right) = \left( \frac{4}{2}, \frac{2}{2}, \frac{7}{2} \right) = (2, 1, 3.5) \] ### Step 2: Calculate the radius \( r \) Calculating the distance from \( M(2, 1, 3.5) \) to \( P(-2, 3, 4) \): \[ r = \sqrt{(2 - (-2))^2 + (1 - 3)^2 + (3.5 - 4)^2} \] Calculating each term: - \( (2 - (-2))^2 = (2 + 2)^2 = 4^2 = 16 \) - \( (1 - 3)^2 = (-2)^2 = 4 \) - \( (3.5 - 4)^2 = (-0.5)^2 = 0.25 \) Now summing these: \[ r^2 = 16 + 4 + 0.25 = 20.25 \] ### Step 3: Write the standard equation of the sphere The center \( M \) is \( (2, 1, 3.5) \) and \( r^2 = 20.25 \). Thus, the equation of the sphere is: \[ (x - 2)^2 + (y - 1)^2 + (z - 3.5)^2 = 20.25 \] This is the standard equation of the sphere. ### Final Answer The equation of the sphere is: \[ (x - 2)^2 + (y - 1)^2 + (z - 3.5)^2 = 20.25 \]

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Bonus Knowledge

To find the equation of the sphere, we first determine the midpoint \( M \) of points \( P(-2, 3, 4) \) and \( Q(6, -1, 3) \). The midpoint \( M \) is calculated as: \[ M = \left( \frac{-2 + 6}{2}, \frac{3 + (-1)}{2}, \frac{4 + 3}{2} \right) = \left( \frac{4}{2}, \frac{2}{2}, \frac{7}{2} \right) = (2, 1, \frac{7}{2}). \] Next, we determine the radius \( r \) of the sphere, which is the distance from the center \( M \) to either point \( P \) or \( Q \). We'll find the distance from \( M \) to \( P \): \[ r = \sqrt{(2 - (-2))^2 + (1 - 3)^2 + \left(\frac{7}{2} - 4\right)^2} = \sqrt{(2 + 2)^2 + (1 - 3)^2 + \left(\frac{7}{2} - \frac{8}{2}\right)^2}. \] Calculating each term: \[ = \sqrt{(4)^2 + (-2)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{16 + 4 + \frac{1}{4}} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{80}{4} + \frac{1}{4}} = \sqrt{\frac{81}{4}}. \] This simplifies to: \[ r = \frac{9}{2}. \] Now, we can write the standard equation of the sphere with center \( M(2, 1, \frac{7}{2}) \) and radius \( r = \frac{9}{2} \): \[ \left(x - 2\right)^2 + \left(y - 1\right)^2 + \left(z - \frac{7}{2}\right)^2 = \left(\frac{9}{2}\right)^2. \] This simplifies to: \[ \left(x - 2\right)^2 + \left(y - 1\right)^2 + \left(z - \frac{7}{2}\right)^2 = \frac{81}{4}. \] Thus, the equation of the sphere is \[ \left(x - 2\right)^2 + \left(y - 1\right)^2 + \left(z - \frac{7}{2}\right)^2 = \frac{81}{4}. \]

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