I found the equation of a spbelow that has actually a center of $(1,-12,8)$ with a radius of 10 and I gained the complying with equation:

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

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

I assumed that it would certainly be this because because it was an xy-airplane, it wouldn"t encompass the z-percentage of the equation.

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



What your answer amounts to is the circle at which the spbelow intersects the aircraft $z=8$.

However before, we"re searching for the intersection of the sphere and also the $x$-$y$ aircraft, offered by $z=0.;$ In various other words, we"re searching for all points of the spright here at which the $z$-component is $0$.

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

Hence the interarea of the sphere and the $x$-$y$ aircraft gives a circle via facility $(1, -12)$ and also radius $6$.


In the $xy$-airplane, $z=0$, yet note that you are looking at $(z-8)^2$ as among the regards to your sphere.


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There is a aircraft provided the equation. If a spright here is tangent to this plane, what is the equation for the sphere?(Spright here Center: $M(4,-2.3)$)

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