Question

Find an equation of sphere passing through the points(0,-2,-4) (2,-1,-1) and having its centre on the straight line 2x-3y=0=5y+2z?

Answer 1

Line is given as intersection of planes:

$$\left\{ \begin{array}{l} 2 x - 3 y = 0 \\ 5 y + 2 z = 0 \end{array} \right. \Rightarrow n _ {1} = (2, - 3, 0), n _ {2} = (0, 5, 2)$$

Then vector on the line is cross product of normales: $n = \begin{vmatrix} i & j & k \\ 2 & -3 & 0 \\ 0 & 5 & 2 \end{vmatrix} = -6i - 4j + 10k = (-6, -4, 10)$

Also vector $(3,2,-5)$ is on the line. If we put $x=3$ in system then $y=2$ , and $z=-5$ . It is point on the line.

Parameterization of the line:

$$\left\{ \begin{array}{l} x = 3 t + 3 \\ y = 2 t + 2 \\ z = - 5 t - 5 \end{array} \right.$$

We have to find some $t$ , that distance to 2 given points will be the same.

$$d _ {1} ^ {2} = (3 t + 3) ^ {2} + (2 t + 2 + 2) ^ {2} + (- 5 t - 5 + 4) ^ {2}$$ $$d _ {2} ^ {2} = (3 t + 3 - 2) ^ {2} + (2 t + 2 + 1) ^ {2} + (- 5 t - 5 + 1) ^ {2}$$ $$d _ {1} ^ {2} = d _ {2} ^ {2}$$ $$\begin{array}{l} (3 t + 3) ^ {2} + (2 t + 3) ^ {2} + 2 (2 t + 3) + 1 + (- 5 t - 4) ^ {2} + 6 (- 5 t - 4) + 9 = \\ = (3 t + 3) ^ {2} - 4 (3 t + 3) + 4 + (2 t + 3) ^ {2} + (- 5 t - 4) ^ {2} \\ \end{array}$$ $$\begin{array}{l} 2 (2 t + 3) + 1 + 6 (- 5 t - 4) + 9 = - 4 (3 t + 3) + 4 \\ 4 t - 3 0 t + 6 + 1 - 2 4 + 9 = - 1 2 t - 1 2 + 4 \\ - 2 6 t - 8 = - 1 2 t - 8 \\ \end{array}$$ $$t = 0 \Rightarrow (x, y, z) = (3, 2, - 5)$$

Distance $d_{1}$ is radius: $r = \sqrt{26}$ .

And equation of sphere is $(x - 3)^2 + (y - 2)^2 + (z + 5)^2 = 26$