Find the standard equation of the sphere that has the point \( (-4,-6,-2) \) and \( (2,5,1) \) as endpoints of a diameter. Center of the Sphere is Radius of the Sphere is Equation of the Sphere is write your answer as a decimal.) Submit Question

To find the standard equation of the sphere with a diameter defined by the points \((-4, -6, -2)\) and \((2, 5, 1)\), we begin by finding the center. The center is obtained by averaging the coordinates of the endpoints: \[ \text{Center} = \left( \frac{-4 + 2}{2}, \frac{-6 + 5}{2}, \frac{-2 + 1}{2} \right) = \left( \frac{-2}{2}, \frac{-1}{2}, \frac{-1}{2} \right) = (-1, -0.5, -0.5) \] Next, we calculate the radius, which is half the distance between the two endpoints. The distance \(d\) between the points is calculated as: \[ d = \sqrt{(2 - (-4))^2 + (5 - (-6))^2 + (1 - (-2))^2} = \sqrt{(6)^2 + (11)^2 + (3)^2} = \sqrt{36 + 121 + 9} = \sqrt{166} \] Thus, the radius \(r\) is: \[ r = \frac{d}{2} = \frac{\sqrt{166}}{2} \] Finally, we can write the equation of the sphere using the standard form \((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\): \[ (x + 1)^2 + \left(y + 0.5\right)^2 + \left(z + 0.5\right)^2 = \left(\frac{\sqrt{166}}{2}\right)^2 \] This simplifies to: \[ (x + 1)^2 + (y + 0.5)^2 + (z + 0.5)^2 = \frac{166}{4} = 41.5 \] So, in summary: Center of the Sphere is: \((-1, -0.5, -0.5)\) Radius of the Sphere is: \(\frac{\sqrt{166}}{2} \approx 6.431\) Equation of the Sphere is: \((x + 1)^2 + (y + 0.5)^2 + (z + 0.5)^2 = 41.5\)