Find the equation of the sphere passing through \( P(-2,3,4) \) and \( Q(6,-1,3) \) with its center at the midpoint of \( P Q \). The standard equation of the sphere is (Simplify your answer.)
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To find the equation of the sphere, we first need to determine the midpoint of the points \( P(-2, 3, 4) \) and \( Q(6, -1, 3) \). The midpoint \( M \) can be calculated using the formula: \[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) \] Substituting the coordinates of \( P \) and \( Q \): \[ M = \left(\frac{-2 + 6}{2}, \frac{3 + (-1)}{2}, \frac{4 + 3}{2}\right) = \left(\frac{4}{2}, \frac{2}{2}, \frac{7}{2}\right) = (2, 1, 3.5) \] Next, we find the radius \( r \) of the sphere, which is the distance from the center \( M(2, 1, 3.5) \) to either point \( P \) or \( Q \). We will calculate the distance from \( M \) to \( P \): \[ r = \sqrt{(x_M - x_P)^2 + (y_M - y_P)^2 + (z_M - z_P)^2} \] Calculating the distance: \[ r = \sqrt{(2 - (-2))^2 + (1 - 3)^2 + (3.5 - 4)^2} = \sqrt{(2 + 2)^2 + (1 - 3)^2 + (3.5 - 4)^2} \] \[ = \sqrt{(4)^2 + (-2)^2 + (-0.5)^2} = \sqrt{16 + 4 + 0.25} = \sqrt{20.25} = \frac{\sqrt{81}}{2} = \frac{9}{2} \] The standard equation of a sphere with center \( (h, k, l) \) and radius \( r \) is given by: \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] Substituting in \( h = 2, k = 1, l = 3.5 \) and \( r = \frac{9}{2} \): \[ (x - 2)^2 + (y - 1)^2 + \left(z - 3.5\right)^2 = \left(\frac{9}{2}\right)^2 \] Calculating \( \left(\frac{9}{2}\right)^2 \): \[ \left(\frac{9}{2}\right)^2 = \frac{81}{4} \] Putting everything together, the equation of the sphere is: \[ (x - 2)^2 + (y - 1)^2 + \left(z - 3.5\right)^2 = \frac{81}{4} \] So, the final simplified equation of the sphere is: \[ (x - 2)^2 + (y - 1)^2 + \left(z - 3.5\right)^2 = \frac{81}{4} \]
