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.
To find the intersection of a sphere with a plane, you need to substitute the equation of the plane into the equation of the sphere and solve for the variable(s) remaining.
In this case, for the intersection with the yz plane, you are correct that the x-coordinate should be zero. The equation of the sphere centered at (2,-10,4) with radius 10 is (x-2)^2 + (y+10)^2 + (z-4)^2 = 100.
Substituting x = 0, we get (0-2)^2 + (y+10)^2 + (z-4)^2 = 100, simplifying to 4 + (y+10)^2 + (z-4)^2 = 100.
To find the intersection with the yz plane, we can set y = 0 in this equation: 4 + (0+10)^2 + (z-4)^2 = 100, which simplifies to (z-4)^2 = 72.
Taking the square root of both sides, we get z-4 = ±√72, or z-4 = ±6√2.
So, the intersection of the sphere with the yz plane consists of two points: (0, -10+6√2, 4) and (0, -10-6√2, 4).
Similarly, for the intersection with the xy plane, you need to set z = 0 in the equation of the sphere: (x-2)^2 + (y+10)^2 + (0-4)^2 = 100.
Simplifying, we have (x-2)^2 + (y+10)^2 + 16 = 100. Setting this equal to zero, we get (x-2)^2 + (y+10)^2 = 84.
So, the intersection of the sphere with the xy plane is described by the equation (x-2)^2 + (y+10)^2 = 84, which represents a circle centered at (2, -10) with radius √84.
To summarize:
- For the intersection with the yz plane, the equation is (z-4)^2 = 72, giving two points: (0, -10+6√2, 4) and (0, -10-6√2, 4).
- For the intersection with the xy plane, the equation is (x-2)^2 + (y+10)^2 = 84, representing a circle centered at (2, -10) with radius √84.
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