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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

Britanica
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  • This could help you: https://math.stackexchange.com/questions/200802/sphere-tangent-to-a-plane – Sebastian Cor Dec 19 '20 at 12:45
  • @SebastianCor But how do I find the center of my sphere? Thanks) – Britanica Dec 19 '20 at 12:49
  • You need the center and the radius to parameterize the sphere. The circle and the sphere will have a line that connects their centers. From the tangent plane you can acquire a normal vector and a point on the sphere which define a line orthogonal to the tangent plane that goes through the center of the sphere. Then you have to find the unique intersection of those two lines to find the center, and then calculate the radius by considering the vector between the center and the point on the tangent plane. – CyclotomicField Dec 19 '20 at 12:51
  • Try a family of spheres that go through your circle, say $x^2+y^2+(z-k)^2=k^2+11,$ and try finding a $k$ that gives tangency. – Jan-Magnus Økland Dec 19 '20 at 12:54
  • @Jan-MagnusØkland OMG! you're a life saver. I knew I have to use one of that family of spheres thing I just didn't know how and I thought I may be wrong. Thnkaaaaank you I can finally go to sleep!!! – Britanica Dec 19 '20 at 12:57
  • My solution is just as @Jan-MagnusØkland says :D – FFjet Dec 19 '20 at 12:59

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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.

FFjet
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