Question

Find the radius of the sphere which passes through the points ( 0,0 ,0 ),(1,0,0 ),(0,1,0)
and (0,0,1)

Answer 1

Question #68628, Math / Other

Find the radius of the sphere which passes through the points $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$

Answer.

Equation of the sphere: $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$.

Points $(0,0,0),(1,0,0),(0,1,0)$ and $(0,0,1)$ lie on the sphere.

So $\left\{ \begin{array}{l}a^{2} + b^{2} + c^{2} = r^{2}\\ (1 - a)^{2} + b^{2} + c^{2} = r^{2}\\ a^{2} + (1 - b)^{2} + c^{2} = r^{2}\\ a^{2} + b^{2} + (1 - c)^{2} = r^{2} \end{array} \right.$ $\rightarrow \left\{ \begin{array}{l}a^{2} + b^{2} + c^{2} = r^{2}\\ 1 - 2a + a^{2} + b^{2} + c^{2} = r^{2}\\ a^{2} + 1 - 2b + b^{2} + c^{2} = r^{2}\\ a^{2} + b^{2} + 1 - 2c + c^{2} = r^{2} \end{array} \right.$ $\rightarrow$

$\left\{ \begin{array}{l}1 - 2a = 0\\ 1 - 2b = 0\rightarrow a = b = c = \frac{1}{2}.\\ 1 - 2c = 0 \end{array} \right.$

Thus, $r^2 = a^2 + b^2 + c^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \rightarrow r = \frac{\sqrt{3}}{2}$.

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