# The equation of a sphere tangent to two planes

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!

• I suppose that the definition of $P$ is $z+3=0$. Commented Feb 9, 2021 at 10:22
• Yes, sorry, that was a typo. Commented Feb 9, 2021 at 10:28

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 $$(x-4)^2+(y-2)^2+(z+4)^2=1$$
• Yes you are write you may get other answers too. One of them is $t=1$. I will see. Commented Feb 9, 2021 at 10:38
• Yest $t=17/7$ is also possible so one more answer. Commented Feb 9, 2021 at 10:50