This is a problem of 3D Geometry. This following question appears in "A Text Book On Co-Ordinate Geometry With Vector Analysis" by Rahman & Bhattacharjee ; Chapter 4 - Exercise 37
If the tangent plane of sphere $\;x^2 + y^2 +(z -2)^2 = 25\;$ cuts the intercepts $\;a,\,b,\,c\;$ from axes prove that,
$25\left(\dfrac1{a^2}+\dfrac1{b^2}+\dfrac1{c^2}\right)=\left(\dfrac2c-1\right)^2$.
I have tried to prove that myself. My progress: I transformed the sphere into a tangent plane form through point P($x_0,y_0,z_0$). Then, the equation of the sphere transforms to, $xx_0 +yy_0 +z(z_0 -2) =21+2z_0$ What should I do next?
I agree with the generic equation of your tangent plane in $x_0 \in S$ :
$$xx_0 +yy_0 +z(z_0 -2)=21+2z_0 \tag{1}$$
We have to keep in mind that $x_0 \in S$ if and only if :
$$x_0^2 + y_0^2 +(z_0 -2)^2 = 25 \tag{2}$$
The intercept $x=a$ of this tangent plane with $x$ axis is obtained by taking $y=z=0$ in (1) giving :
$$a=\frac{21+2z_0}{x_0}$$
Taking now $x=z=0$ in (1), we get the ordinate of the intercept of the plane with $y$ axis :
$$b=\frac{21+2z_0}{y_0}$$
For a similar reason :
$$c=\frac{21+2z_0}{z_0-2}$$
Now, if we compute the left hand side of the relationship you have to establish, we get :
$$25 \left(\frac{x_0^2 + y_0^2 +(z_0 -2)^2}{(21+2z_0)^2}\right)=25 \left(\frac{25}{(21+2z_0)^2}\right)$$
the second expression coming from (2).
which is indeed equal to the right hand side,
$$\left(\dfrac{2(z_0-1)}{21+2z_0}-1\right)^2$$
as you can check it.