Find the pt of tangency of 2 spheres: (x-6)^2+(y+2)^2+(z-4)^2-19=0 and x^2+y^2-4x+6y-72=0?

Answer 1

The point is nowhere in the world.

Answer 2

First way:use your imagination. Try to dram a map or something

Second way, do it:
1: x^2+Y^2-12x+4y+(z-4)^2=-21
2:x^2+y^2-4x+6y=72
2-1:16x+2y-(z-4)^2=93 So, (z-4)^2=16x+2y-93
On the other hand, (x-2)^2+(y+3)^2=85=81+4=49+36.
I don't have a calculator at hand now, so I can only do it this way:
Possible 1: x-2=9 and y+3=2, then y<0, out.
Possible 2:x-2=2 and y+3=9, then it's impossible. out.
Possible 3:x-2=7and y+3=6, then x=9 and y=3, then (z-4)^2=57, it's possible.
Possible 4:x-2=6and y+3=7,then x=8and y=4,then (z-4)^2=43, possible.

Then find a calculator and find the answer.

Answer 3

The second equation is not a sphere. It has no z values.

Please check your question and resubmit.