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.
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.
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)