Answer to Question #166804 in Analytic Geometry for Anshika

Question

Tangent plane at any point of sphere x^2+y^2+z^2=r^2 meets the coordinate axis at A,B, C. Show that the locus of the point of intersection of planes drawn parallel to the coordinate planes through A,B,C is the surface x^-2+y^-2+z^-2=r^-2

Answer 1

The equation of the tangent plane at any point $(x_0,y_0,z_0)$ :

$xx_0+yy_0+zz_0=r^2$

The plane meets the coordinate axes at

$A:(r^2/x_0,0,0),B:(0,r^2/y_0,0),C:(0,0,r^2/z_0)$

The equations of the planes parallel to the coordinate planes through A, B, C:

$x=r^2/x_0,y=r^2/y_0,z=r^2/z_0$

Then:

$x^{-2}+y^{-2}+z^{-2}=\frac{1}{r^2}(x_0^2+y_0^2+z_0^2)=\frac{r^2}{r^4}$

$x^{-2}+y^{-2}+z^{-2}=r^{-2}$