the equation of the sphere is
(x-2)^2 + (y+10)^2 + (z-4)^2 = 100
so in the xz plane: y = 0
(x-2)^2 + 100 + (z-4)^2 = 100
(x-2)^2 + (x-4)^2 = 0
conclusion: the yz plane just touches the sphere.
for xy plane , let z=0
(x-2)^2 + (y+10)^2 + 16 = 100
(x-2)^2 + (y+10)^2 = 84
Visualize the sphere having a slice cut off, and you are looking at the surface of the cut.
Wouldn't that be a circle?
And unless the cut is made through the middle of the sphere, wouldn't that circle have a radius less than 10 ?
So for my answer of (x-2)^2 + (y+10)^2 = 84, the centre of that surface would be (2, -10,0) and the radius of the cut surface would be √84
you had the right idea, but you just dropped the whole bracketted term, you should have just let the y=0
First of all let me apologize if this is the wrong section; this is a review question from a Calculus III class, but it doesn't involve any actual calculus so I posted it here.
If I have a sphere with center (2,-10,4) and radius 10 what is:
the intersection with the yz plane?
the intersection with the xy plane?
I get tripped up with geometry in more than 2 dimensions so I'm confused as how to handle this.
Should I treat the "omitted plane" (my term for lack of a better word) value as zero? If so does that change the radius? I'm guessing not given that the distance formula for a circle or a sphere is equal to r^2.
So for yz would I say
(y+10)^2 + (z-4)^2 = 100
And if so, how do I find the yz plane intersection from that? I know I've done this before but I don't recall how to do it.
1 answer
1 answer