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.
Best Answer
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.