I uncovered the equation that a round that has actually a center of \$(1,-12,8)\$ through a radius the 10 and also I obtained the complying with equation:

\$(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. Thanks because that contributing an answer to doyourpartparks.org Stack Exchange!

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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)\$) site architecture / logo © 2021 stack Exchange Inc; user contributions licensed under cc by-sa. Rev2021.11.25.40831