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I got a maths question that gives you 3 points, $A (6,0,0)$ and $B (6,6,0)$ and $C (0,6,0)$ and a line DG, D being $(0,0,6)$ and $G (0,6,6)$ so the equation of DG is $\vec r$ = 6$\hat i$ -$6t\hat j$ . You're then asked to find the equation of the sphere that contains those 3 points and is tangential to the line.

I know that if I had 4 points I could do a simultaneous equation(s), but I don't see how that could work here. Please help.

I should maybe mention that the $origin, A, B, C, D, G, E(6,0,6) ~and~ F (6,6,6)$ form the 3 corners of a cube.

Sebastian
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A sphere can be represented as $$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$ where the center is $(a,b,c)$ and its radius is $r$.

Setting each of $(x,y,z)=(6,0,0),(6,6,0),(0,6,0)$ in the equation, you'll get $$(6-a)^2+(0-b)^2+(0-c)^2=r^2,$$ $$(6-a)^2+(6-b)^2+(0-c)^2=r^.2$$ $$(0-a)^2+(6-b)^2+(0-c)^2=r^2.$$

Hence, you'll get $$(a,b,c)=(3,3,r^2-18).$$

Now we know that the equation of the sphere is represented as $$(x-3)^2+(y-3)^2+(z-r^2+18)^2=r^2.$$

Then, you can set $$(x,y,z)=(0, 6t, 6)$$ in the equation (because the line $DG$ is represented as $(0,0,6)+t(0,6,0)$.)

Then, you'll have an equation with $t,r$, so what you do is to find $r$ such that the equation has only one real solution $t$.

mathlove
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  • One can also use the fact that the distance of the co-ordinate of the center of the sphere from the tangent line has to be equal to r. – GTX OC Jan 05 '14 at 12:26
  • "Setting three points in the equation, you'll get (a,b,c)=(3,3,r2−18). Now we know that the equation of the sphere is represented as (x−3)2+(y−3)2+(z−r2+18)2=r2."

    Wait, how did you get the coordinates of the centre of the sphere?

     – Sebastian Jan 05 '14 at 12:45
  • @Sebastian: I added a bit. see my answer. – mathlove Jan 05 '14 at 14:23