Sphere centered in line $(x,y,z) = (-2,0,1) + \lambda (0,0,1)$ tangent to planes $ x-10z = 0 $ and $ x+2z = 0 $ whose radius squared is $r^2 > 20$

Question

how may I find the sphere centered in line

$$(x,y,z) = (-2,0,1) + \lambda (0,0,1)$$

tangent to planes $$ x-10z = 0 $$

and

$$ x+2z = 0 $$

whose radius squared is

$$r^2 > 20$$

Thank you.

@Dr.SonnhardGraubner hello Dr. Sonnhard, I just attached a screen capture from Geogebra of the problem. Can you help me figure it out? — bru1987, Nov 29 '15 at 13:55

score 1 · Accepted Answer · answered Nov 29 '15 at 14:04

Hint:

Note that the problem can be reduced to the plane $x-z$ (i.e. $y=0$) where the section of the sphere is a circle of center $C=(-2,\lambda)$ and the sections of the two planes are the straight lines: $$ \alpha) \quad x-10z=0 \qquad \beta) \quad x+2z=0 $$

The center have to be such that its distances from these lines are the same: $$ \frac{|-2-10\lambda|}{\sqrt{1^2+10^2}}=\frac{|-2+2\lambda|}{\sqrt{1^2+2^2}} $$

can you do from this?

Yes I can Emilio, thank you very much. Best Regards. – bru1987 Nov 29 '15 at 14:16 — bru1987, Nov 29 '15 at 14:16

Sphere centered in line $(x,y,z) = (-2,0,1) + \lambda (0,0,1)$ tangent to planes $ x-10z = 0 $ and $ x+2z = 0 $ whose radius squared is $r^2 > 20$

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