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The center of a sphere belongs to the line $$d : {{x-1}\over3} = {{y}\over 2} = {{z+2}\over-2} $$ Find the equation of the sphere if the planes $$ (P):z+3=0$$ $$ (Q) :3x-4z-33=0 $$ are tangent to the sphere.

My solution: from the parametric equations of the line, I found that the center of the sphere is of the form $C(1+3t,2t,-2-2t)$. Now, if the planes are tangent to the sphere then the distance from the center of the sphere to each plane is equal with the radius. The thing is, I've computed both of these but I'm stuck at a point where I have the absolute value of two expressions. Would it also be viable to find the line as the intersection of the two planes and work with that? Thank you, in advance!

Andrei0408
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Perpendicular to the first plane from center is $d_1=-2-2t+3=0=1-2t$ Similarly $d_2=(17t-22)/5$ Equate $d_1=d_2$ to get $t=1$, So now you know the center $(4,2,-4)$ and the radius $r=|d_1|=|d_2|=1.$ to write the Eq. of sphereas $$(x-4)^2+(y-2)^2+(z+4)^2=1$$

Z Ahmed
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  • d1 and d2 are the distances from the center to each plane right? because if so, I've gotten the same values except for d2, for me it was (17t-22)/5 and from this t = 1. But I have a question, if these are the distances then the equality is |1-2t| = |17t-22| / 5 (with absolute value), how do I get rid of the absolute value if I have a variable in both expressions (it seems like it would take a lot of computation, 2 cases for each expression). – Andrei0408 Feb 09 '21 at 10:35
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    Yes you are write you may get other answers too. One of them is $t=1$. I will see. – Z Ahmed Feb 09 '21 at 10:38
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    Yest $t=17/7$ is also possible so one more answer. – Z Ahmed Feb 09 '21 at 10:50