Question

Find the equation of a sphere if one of its diameters has endpoints: (-9, -12, -6) and (11, 8, 14).
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Answer 1

Answer:

Hence, the equation of a sphere with one of its diameters with endpoints (-9, -12, -6) and (11, 8, 14) is $(x-1)^{2}+(y+2)^{2}+(z-4)^{2} = 30$.

Step-by-step explanation:

There are two kew parameters for a sphere: Center ($h$, $k$, $s$) and Radius ($r$). The radius is the midpoint of the line segment between endpoints. That is:

$C(x,y,z) = \left(\frac{-9+11}{2},\frac{-12+8}{2},\frac{-6+14}{2} \right)$

$C(x,y,z) = (1,-2,4)$

The radius can be found by halving the length of diameter, which can be determined by knowning location of endpoints and using Pythagorean Theorem:

$r = \frac{1}{2}\cdot \sqrt{(-9-11)^{2}+(-12-8)^{2}+(-6-14)^{2}}$

$r = 10\sqrt{3}$

The general formula of a sphere centered at (h, k, s) and with a radius r is:

$(x-h)^{2}+(y-k)^{2}+(z-s)^{2} = r^{2}$

Hence, the equation of a sphere with one of its diameters with endpoints (-9, -12, -6) and (11, 8, 14) is $(x-1)^{2}+(y+2)^{2}+(z-4)^{2} = 30$.