$(x-1)^2 + (y+12)^2 + (z-8)^2 = 100$

As for finding an intersection for the xy-plane I obtained this and also it was the dorn answer:$(x-1)^2 + (y+12)^2 = 100$

I assumed that it would certainly be this due to the fact that since it to be an xy-plane, the wouldn"t incorporate the z-portion that the equation.

You are watching: What is the intersection of this sphere with the yz-plane?

What your answer amounts to is the circle in ~ which the sphere intersects the aircraft $z=8$.

However, we"re trying to find the intersection of the sphere and the $x$-$y$ plane, provided by $z=0.;$ In various other words, we"re in search of all point out of the round at which the $z$-component is $0$.

Thus we should evaluate the round using $z = 0,,$ which returns the circle $$eginalign (x-1)^2 + (y+12)^2 + (0-8)^2 = 100 &iff (x-1)^2 + (y+12)^2 + 64 = 100\ \ &iff (x-1)^2 + (y+12)^2 = 36 = 6^2endalign$$

Thus the intersection the the sphere and the $x$-$y$ plane gives a one with facility $(1, -12)$ and radius $6$.

In the $xy$-plane, $z=0$, yet note the you room looking in ~ $(z-8)^2$ as one of the terms of your sphere.

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over there is a plane given the equation. If a ball is tangent come this plane, what is the equation because that the sphere?(Sphere Center: $M(4,-2.3)$)

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