find an equation of the sphere with center (2,-6,4) and radius 5. Describe its intersection with each of the coordinates planes.

- anonymous

- chestercat

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- anonymous

so i know how to do the equation, but how do i do the intersection?

- anonymous

\[(x-2)^2 + (y+6)^2 + (z-4)^2 = 25\]

- Hero

What course are you taking?

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## More answers

- anonymous

calc 3

- anonymous

the description of the intersection is an equation of a circle i believe based on the example prior to this problem

- amistre64

|dw:1327252360580:dw|

- amistre64

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

- amistre64

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)

- amistre64

the distance away from the plane is the one that you dropped :)

- anonymous

okay so the xy plane would be (x-2)^2 + *y+6)^2 = 9

- anonymous

The sphere does not intersect the y=0 coordinate plane.

- anonymous

wait it would be = 16?

- amistre64

(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)

- anonymous

ohhh okay okay

- anonymous

wait still confused

- anonymous

so what would be the equation for yz?

- Hero

How do I unsubscribe from this?

- amistre64

yz: no-x plane; centers on (y,z) with a radius of: sqrt(5^2 - x^2)

- anonymous

so 25-4? root 21?

- amistre64

yep

- amistre64

and r^2 = 21 then in your equation of circle stuff

- anonymous

but how would i do xz?

- anonymous

r = root(25-36)

- amistre64

well, it centers in the no-y so (x,z); its radius is then sqrt(5^2-y^2)

- amistre64

yep

- anonymous

but its a negative number

- amistre64

it is isnt it :) tell me; if your stretching to reach 5 and your 6 away, do you hit it?

- anonymous

yeah bc r =6?

- amistre64

|dw:1327252967020:dw|

- anonymous

oh so it never intersects?

- amistre64

its negative meaning there is no real radius in that plane so we aint even touching it

- anonymous

so there wouldn't be an equation for xz

- amistre64

correct; the xz plane has no intersection with the sphere

- anonymous

awesome! thank you so much!

- amistre64

yep

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