I know my equation is $(x+3)^2+(y-2)^2+(z-5)^2=16$ but how do I find the intersection with the y-z plane?
I'm doing some study prep for my upcoming calculus 3 class.
2026-03-30 08:57:37.1774861057
Find an equation of the sphere with center $(-3,2,5)$ and radius $4$. What is the intersection of this sphere with the y-z plane?
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The $yz$-plane is the plane $x=0.$ So set $x=0$ to your equation and you will get $(y-2)^2 +(z-5)^2=16-9=7$ which is a circle on the $yz$-plane.