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.

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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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