My problem is the following: Find the equation of the sphere, that passes through the circle $$x^2+y^2=11$$, $$z=0$$ and is tangent to the plane $$x+y+z-5=0$$. I have no idea how to solve this one. I know this looks easy, but I guess I'm stupid and I have an analytic geometry exam in a week. I would really appreciate any hint or solution. Thanks in advance!
P.S.Please don't bully me for asking stupid questions. I'm a sensitive person
Assume the center of the sphere is $(0,0,t)$, then the equation of the sphere is $$ x^2+y^2+(z-t)^2=t^2+11 $$ Using the formula of Distance from point to plane, we have $$ d= \frac{1\cdot 0+1\cdot 0+1\cdot t-5}{\sqrt{1^2+1^2+1^2}} = \dfrac{t-5}{\sqrt{3}} $$ Since it is tangent to the plane, the following relation holds: $$ d = r = \sqrt{t^2+11} $$ Simplify, and we get $$ \left( \frac{t-5}{\sqrt{3}} \right) ^2=t^2+11\\ \dfrac{t^2-10t+25}{3}=t^2+11\\ t^2-10t+25=3t^2+33 \\ 2t^2+8=-10t \\ t^2+5t+4=0 $$ Thus $$ t=-1 \text{ or } t = -4 $$ We are done.