Consider the intersection of the sphere (x-3)^2+(y+2)^2+(z-1)^2=13 with the plane x+y=0(adsbygoogle = window.adsbygoogle || []).push({});

a) This intersection should be a familiar curve. Describe the curve.

b) Substituing x=-y or y=-x into the equation of the sphere does not give this curve. Explain the difference.

I know that the intersectionis a circle but I am having trouble proving that. I rearranged the above equation and set it equal to x+y:

(x-3)^2+(y+2)^2+(z-1)^2-13=x+y

The next thing I did was put the like terms together to give the following equation:

x^2-6x+9-x+y^2+4y+4+y+z^2-2z+1-13=0

I then simplified this equation:

X^2-7x+9+y^2+3y+4+(z-1)^2-13=0

Now I am stuck. This just looks like a smaller sphere.

Also, I would plug in x=-y or vice versa but the second part of the question says that will not give me the result that I want.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Calc III: sphere & plane intersection.

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