Question

a) Find the equation of the sphere which passes through the points
(1, − 4, 3), (1, − 5, 2), (1, − 3, 0) and whose centre lies on the plane x + y + z = 0 .

Answer 1

Answer on Question #65190 – Math – Analytic Geometry

Question

Find the equation of the sphere which passes through the points $(1, -4, 3)$, $(1, -5, 2)$, $(1, -3, 0)$ and whose centre lies on the plane $x + y + z = 0$

Solution

Mentioned points define some plane. Cross-section of plane and sphere is circle. As all points have the same $x$-coordinate, then circle lay on $yz$-plane. Hence, we can write

$$\left\{ \begin{array}{c} (-4 - y_0)^2 + (3 - z_0)^2 = r^2, \\ (-5 - y_0)^2 + (2 - z_0)^2 = r^2, \quad \rightarrow \quad \left\{ \begin{array}{c} (y_0 + 4)^2 + (z_0 - 3)^2 = r^2, \\ (y_0 + 5)^2 + (z_0 - 2)^2 = r^2, \\ (y_0 + 3)^2 + z_0^2 = r^2, \end{array} \right. \\ \end{array} \right.$$

where $y_0, z_0$ are coordinates of the center of the circle and, at the same time, coordinates of the center of the sphere; $r$ is a radius of the circle.

Solve system for $y_0$ and $z_0$:

$$\left\{ \begin{array}{l} (y_0 + 4)^2 + (z_0 - 3)^2 = (y_0 + 3)^2 + z_0^2 \\ (y_0 + 5)^2 + (z_0 - 2)^2 = (y_0 + 3)^2 + z_0^2 \end{array} \right.$$ $$\left\{ \begin{array}{l} (2y_0 + 7) - 3(2z_0 - 3) = 0 \quad \rightarrow \quad \left\{ \begin{array}{l} 2y_0 - 6z_0 + 16 = 0 \\ 4y_0 - 4z_0 + 20 = 0 \end{array} \right. \quad \rightarrow \quad \left\{ \begin{array}{l} y_0 - 3z_0 + 8 = 0 \\ y_0 - z_0 + 5 = 0 \end{array} \right. \end{array} \right.$$ $$2z_0 - 3 = 0 \rightarrow z_0 = \frac{3}{2}. \text{ Then } y_0 = z_0 - 5 \rightarrow y_0 = -\frac{7}{2}$$

$x_0$-coordinate can be found from the following condition:

$$x_0 + y_0 + z_0 = 0 \rightarrow x_0 = -y_0 - z_0$$ $$x_0 = \frac{7}{2} - \frac{3}{2} = \frac{4}{2} = 2$$

Hence the center of the sphere is

$$\left(2, -\frac{7}{2}, \frac{3}{2}\right).$$

The only thing left is to find squared radius of the sphere. Let's calculate the distance from the center to the $1^{\text{st}}$ point:

$$R^2 = (1 - 2)^2 + \left(-4 - \left(-\frac{7}{2}\right)\right)^2 + \left(3 - \frac{3}{2}\right)^2 \rightarrow$$ $$R^2 = (-1)^2 + \left(-\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2 \rightarrow R^2 = 1 + \frac{1}{4} + \frac{9}{4} = \frac{14}{4}$$

Thus, equation of the sphere is

$$(x - 2)^2 + \left(y + \frac{7}{2}\right)^2 + \left(z - \frac{3}{2}\right)^2 = \frac{14}{4}.$$

Answer: $(x - 2)^2 + \left(y + \frac{7}{2}\right)^2 + \left(z - \frac{3}{2}\right)^2 = \frac{14}{4}$.

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