Answer to Question #114947 in Analytic Geometry for ANJU JAYACHANDRAN

Question

A plane passes through (a,b, c) and cuts the axes in A,B,C, respectively, where
none of these points lie on the origin O. Show that the centre of the sphere OABC
satisfies the equation a/x +b/y +c/z = 2.

Answer 1

Let the equation of plane be $kx+ly+mz+n=0.$ The plane passes through (a,b,c), so

$ka+lb+mc+n=0.$ (1)

We may calculate A, B, C as

$kA+n = 0 \Rightarrow A = -\dfrac{n}{k}, k = - \dfrac{n}{A},\\ lB+n = 0 \Rightarrow B = -\dfrac{n}{l}, l = - \dfrac{n}{B}, \;\;\;\;\;\;(2)\\ mC+n = 0 \Rightarrow C = -\dfrac{n}{m}, m = - \dfrac{n}{C}.$

If O, A, B, C are situated on the sphere with radius $r$ and center in $(x_0,y_0,z_0),$ then

$x_0^2 + y_0^2 + z_0^2 = r^2, \;\; (x_0+\frac{n}{k})^2 + y_0^2 + z_0^2 = r^2, \\ x_0^2 + (y_0+\frac{n}{l})^2 + z_0^2 = r^2, \;\; x_0^2 + y_0^2 + (z_0+\frac{n}{m})^2 = r^2.$

Therefore, $2x_0\frac{n}{k} + \left(\frac{n}{k} \right)^2 = 0,$ so $x_0 = \frac{A}{2}.$ Similarly we obtain $y_0 = \frac{B}{2}, \;\; z_0 = \frac{C}{2}. \;\;\; (3)$

From (2) and (1) we get

$-\dfrac{n}{A} a -\dfrac{n}{B} b-\dfrac{n}{C} c + n = 0,$ therefore $\dfrac{a}{A} + \dfrac{b}{B} + \dfrac{c}{C} = 1.$

Next we substitute expressions from (3):

$\dfrac{a}{2x_0} + \dfrac{b}{2y_0} + \dfrac{c}{2z_0} = 1, \\ \dfrac{a}{x_0} + \dfrac{b}{y_0} + \dfrac{c}{z_0} = 2.$