The number of common tangent planes to the spheres $(x+2)^2+y^2+z^2=1 , (x-2)^2+y^2+z^2=1$ passing through the origin is equal to
(A) 0 (B) 1 (C) 2 (D) None of these .
My work :
Drawing the graph it seems the spheres do not touch each other . Hence has no common tangent . This is also follows from the equation : Solving the two equations we get $x=0$ . So if the spheres are touching each other at the point $(o,\alpha , \beta) $ then $(0-2)^2+\alpha^2+\beta^2=1 $ which clearly does not have any real solution .
Hence the answer must be (A) $0$ .
Is my solution correct ?
There are infinitely many lines which are tangent to one of the sphere $S_1$ and pass through the origin. By symmetry, such lines will also be tangent to the another sphere $S_2$.
Let $l$ be one of such line and that it touches $S_1$ at point $p$. Consider the tangent plane $P$ of $S_1$ at $p$. Then $P$ contains $l$ and is tangent to $S_2$ by symmetry.
Hence there should be infinitely many of them and the answer is (D).