Determine constant $\alpha$ such that tangent planes on given spheres in every intersection point are orthogonal

Question

$\pi_1..... x^2+(y-\alpha)^2+z^2=3$

$\pi_2..... (x-1)^2+y^2+z^2=1 $

Tangent planes in $P_0=(x_0,y_0,z_0) \in \pi_1, \pi_2$

$T_1....2x_0(x-x_0)+2(y_0-\alpha)(y-y_0)+2z_0(z-z_0)=0$

$T_2....2(x_0-1)(x_0-x)+2y_0(y-y_0)+2z_0(z-z_0)=0$

I think I need to get system of 3 linear equations with 3 unknowns. Where do I go from here?

EDIT: I am not sure if I am on right path but

$T_1....2 x_0 x+2(y_0-\alpha)y+2z_0z= L_1$

$T_2....2(x_0-1)x_0+2y_0y+2z_0z= L_2$

They are perpendicular if $n_1 n_2=0 \iff 2x_02(x_0-1)+2(y_0-\alpha)2y_0+4z_0^2=0$

$$x^2+(y-\alpha)^2+z^2=3 $$ $$ (x-1)^2+y^2+z^2=1 $$ $$ x(x-1)+(y-\alpha)y+z^2=0$$

Answer 1

Let $C_i$ be the center of sphere $\pi_i$, with $C_1=(0,\alpha,0)$, $C_2=(1,0,0)$ and the respective radii $\sqrt{3}$ and $1$.

First of all, it is clear, because the problem has a rotational invariance around the center line $C_1C_2$, that it suffices to have a reasoning on a single common point $P_0$.

In cross-sectional plane $P_0C_1C_2$, we have this 2D figure with circles the same radii as the corresponding spheres and a right angle $\widehat{C_1P_0C_2}$:

Explanation about the right angle: $C_1P_0$ being orthogonal to the tangent plane to $\pi_1$ in $P_0$, and $C_2P_0$ being orthogonal to the tangent plane to $\pi_2$ in $P_0$, we deduce that $C_1P_0 \perp C_2P_0$.

Let us write Pythagoras' theorem:

$$\tag{1}(C_1C_2)^2=(C_1P_0)^2+(C_2P_0)^2.$$

with $C_1P_0=\sqrt{3}$ (radius of $\pi_1$), $C_2P_0=1$ (radius of $\pi_2$) and distance $C_1C_2=\sqrt{(1-0)^2+(\alpha-0)^2+(0-0)^2}= \sqrt{1+\alpha^2} \ .$ Thus (1) gives:

$$1+\alpha^2=\sqrt{3}^2+1^2=4$$ yielding the 2 values

$\alpha = \pm \sqrt{3}$

for which the spheres are orthogonal.