Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$ Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$

My Attempt:
Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.As this sphere touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$.
So $1+4+4+2u+4v-4w+d=0\implies2u+4v-4w+d=-9.......(1)$
Also the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ passes through $(1,-1,0)$
So $1+1+2u-2v+d=0\implies2u-2v+d=-2......(2)$
I am stuck here.Please help me.
 A: Hints:
(1) The given sphere is $\;(x+1)^2+(y-3)^2+z^2=9\;$
(2) The line through the above sphere's center and the given tangency point is $\;(-1,3,0)+t(1,2,-2)\;,\;\;t\in\Bbb R\;$
(3) In a similar way as in plane geometry, two tangent spheres' centers are joined by a line passing through the tangency point.
A: As sphere touch each other at $(1,2,-2)$ so distance between their centres=$r_1+r_2$ and distance between centre of sphere and point $(1,-1,0)$ is the radius of sphere(at that point we have a tangent). Now you have $4$ equations and $4$ unknowns you can now solve them
A: You're off to a good start. You've used the fact that your sphere must pass through $(1,2,-2)$, but not that it touches the other sphere in this point. Your sphere touches the other sphere in this point if and only if their tangent planes are the same in this point. This gives you a third equation for $u$, $v$, $w$ and $d$.
A: I have tried my best to answer the question if any doubt arises then post.
But i think its qite easy to understand provided you have the knowledge of directio ratios and cosine.
Thanks.
Happy learning
(https://i.stack.imgur.com/OXuSf.jpg)
