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so i know how to do the equation, but how do i do the intersection?
\[(x-2)^2 + (y+6)^2 + (z-4)^2 = 25\]
What course are you taking?
the description of the intersection is an equation of a circle i believe based on the example prior to this problem
you know the distance you are from the plane; you know the Radius of the sphere, so we can calculate the radius of the circle in the plane
the plane you pick is just dropping that variable from the picture to get your center in that plane: (1,2,3) centers in no-x plane at (2,3) the no-y plane at (1,3) and the no-z plane at (1,2)
the distance away from the plane is the one that you dropped :)
okay so the xy plane would be (x-2)^2 + *y+6)^2 = 9
The sphere does not intersect the y=0 coordinate plane.
wait it would be = 16?
(2,-6,4) and radius 5 xy: no-z: center at (2,-6) with a distance of 4 radius in no-z = sqrt(5^2-4^2)
ohhh okay okay
wait still confused
so what would be the equation for yz?
How do I unsubscribe from this?
yz: no-x plane; centers on (y,z) with a radius of: sqrt(5^2 - x^2)
so 25-4? root 21?
and r^2 = 21 then in your equation of circle stuff
but how would i do xz?
r = root(25-36)
well, it centers in the no-y so (x,z); its radius is then sqrt(5^2-y^2)
but its a negative number
it is isnt it :) tell me; if your stretching to reach 5 and your 6 away, do you hit it?
yeah bc r =6?
oh so it never intersects?
its negative meaning there is no real radius in that plane so we aint even touching it
so there wouldn't be an equation for xz
correct; the xz plane has no intersection with the sphere
awesome! thank you so much!