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ag
Numerade Educator

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Problem 13 Easy Difficulty

Find an equation of the sphere with center $ (-3, 2, 5) $ and radius 4. What is the intersection of this sphere with the $ yz $-plane?

Answer

$(x+3)^{2}+(y-2)^{2}+(z-5)^{2}=16$:$(y-2)^{2}+(z-5)^{2}=7, x=0$ (a circle)

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Video Transcript

So we know that the equation of a sphere experience eight square. Why might this case squared? Plus the minus a many's a letter end. You know what your book uses, but it doesn't really matter. So the only plug in our center? Yeah, and sixteen. And then we want to see where intersects. We want to see where intersects the y Z plane. So the wise e plan has an exploding it zero. So if we plug in zero, we're going get three square. That's why minus two squared lysine minus five. Where is equal to sixteen, and then we end up with concert. What is something that looks like this? So this is a circle center is to five in the radius is rad seven we can figure out that's that is the intersection.