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Jet001 [13]
3 years ago
14

Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i

ts intersection with each of the coordinate planes. (if the sphere does not intersect with the plane, enter dne.)
Mathematics
1 answer:
lana66690 [7]3 years ago
3 0

In the x,y plane, we have z=0 everywhere. So in the equation of the sphere, we have

25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2

which is a circle centered at (2, -10, 0) of radius 4.

In the x,z plane, we have y=0, which gives

25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the y,z plane, x=0, so

25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2

which is a circle centered at (0, -10, 3) of radius \sqrt{21}.

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f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

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