Question

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

Answer 1

Answer on Question #62265 – Math – Analytic Geometry

Question

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

Solution

Let $x, y, z$ be coordinates of the center of the sphere.

$$\begin{array}{l} \left\{ \begin{array}{l} \sqrt{(x - 0)^2 + (y - 3)^2 + (z - 1)^2} = \sqrt{(x - 2)^2 + (y - 5)^2 + (z - 1)^2} = \sqrt{(x - 3)^2 + (y - 4)^2 + (z - 1)^2} \\ x + y + z = 0 \end{array} \right. \Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} x^2 + (y - 3)^2 + (z - 1)^2 = (x - 2)^2 + (y - 5)^2 + (z - 1)^2 \\ x^2 + (y - 3)^2 + (z - 1)^2 = (x - 3)^2 + (y - 4)^2 + (z - 1)^2 \end{array} \right. \Rightarrow \\ x + y + z = 0 \\ \Rightarrow \left\{ \begin{array}{l} x^2 + y^2 - 6y + 9 + z^2 - 2z + 1 = x^2 - 4x + 4 + y^2 - 10y + 25 + z^2 - 2z + 1 \\ x^2 + y^2 - 6y + 9 + z^2 - 2z + 1 = x^2 - 6x + 9 + y^2 - 8y + 16 + z^2 - 2z + 1 \end{array} \right. \Rightarrow \\ x + y + z = 0 \\ \Rightarrow \left\{ \begin{array}{l} -6y - 2z + 10 = -4x - 10y - 2z + 30 \\ -6y - 2z + 10 = -6x - 8y - 2z + 26 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 4x + 4y = 20 \\ 6x + 2y = 16 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = 5 - x \\ 6x + 10 - 2x = 16 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = 5 - x \\ 4x = 6 \\ z = -5 \end{array} \right. \\ \Rightarrow \left\{ \begin{array}{l} x = 1.5 \\ y = 3.5 \\ z = -5 \end{array} \right. \end{array}$$

We found the center of the sphere is $(1.5, 3.5, -5)$.

The radius of the sphere is

$$r = \sqrt{(1, 5 - 0)^2 + (3, 5 - 3)^2 + (-5 - 1)^2} = \sqrt{2, 25 + 0, 25 + 36} = \sqrt{38.5} \Rightarrow r^2 = 38.5.$$

The equation of the sphere is

$$(x - 1.5)^2 + (y - 3.5)^2 + (z + 5)^2 = 38.5.$$

Answer: $(x - 1.5)^2 + (y - 3.5)^2 + (z + 5)^2 = 38.5$

www.AssignmentExpert.com