Sphere

Given the sphere $\mathbb{S}: x^2 +y^2 +z^2 -2x + 2y -4z +2 =0$ and the plane $(\pi): 3x-2y +z =5$ show that:

a) the two surfaces intersect.
b) The radius of the circle formed by the intersection is $\displaystyle R = \frac{26}{7}$. In continuity evaluate its center.

Solution



a) The shpere takes its equivelant form $\mathbb{S}: (x-1)^2 + (y+1)^2 +(z-2)^2 =4$. This , in return , means that the center of the sphere is $K_1 (1, -1, 2)$ and its radius is $R_1 =2$.Now it is a routine process to see that these two surfaces indeed intersect.

b) The circle formed by the intersection has an equation of:

$$\left\{\begin{matrix}
(x-1)^2 + (y+1)^2 + (z-2)^2 =4  \\
 3x-2y +z =5 \tag {1}
\end{matrix}\right.$$

The center $K_2$ of the circle will be found by intersecting the perpendicular line passing through $K_1$. This line has an equation

$$(\varepsilon):  \frac{x-1}{3}=\frac{y+1}{-2}= \frac{z-2}{1} \tag{2} $$

since it is parellel to the vector $\vec{\eta} (3, -2, 1)$ (that is the vector that is perpendicular to the plane).

Solving the system defined by $(1), (2)$ (where $(1)$ is the equation of the plane) we get the coordinates of the center of the circle, that is $$ K_2 \left( \frac{4}{7}, \; -\frac{5}{7}, \; \frac{13}{7} \right)$$

To evaluate the radius we apply Pythagora's theorem. Hence
$$R=\sqrt{R_1^2 - \left | \overrightarrow{K_1K_2} \right |^2}= \frac{26}{7}$$

and we get the wanted radius.