<span>Sphere: (x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
Intersection in xy-plane: (x - 4)^2 + (y + 12)^2 = 36
Intersection in xz-plane: DNE
Intersection in yz-plane: (y + 12)^2 + (z - 8)^2 = 84
The desired equation is quite simple. Let's first create an equation for the sphere centered at the origin:
x^2 + y^2 + z^2 = 10^2
Now let's translate that sphere to the desired center (4, -12, 8). To do that, just subtract the center coordinate from the x, y, and z variables. So
(x - 4)^2 + (y - -12)^2 + (z - 8)^2 = 10^2
(x - 4)^2 + (y - -12)^2 + (z - 8)^2 = 100
Might as well deal with that double negative for y, so
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
And we have the desired equation.
Now for dealing with the coordinate planes. Basically, for each coordinate plane, simply set the coordinate value to 0 for the axis that's not in the desired plane. So for the xy-plane, set the z value to 0 and simplify. So let's do that for each plane:
xy-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + (y + 12)^2 + (0 - 8)^2 = 100
(x - 4)^2 + (y + 12)^2 + (-8)^2 = 100
(x - 4)^2 + (y + 12)^2 + 64 = 100
(x - 4)^2 + (y + 12)^2 = 36
xz-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + (0 + 12)^2 + (z - 8)^2 = 100
(x - 4)^2 + 12^2 + (z - 8)^2 = 100
(x - 4)^2 + 144 + (z - 8)^2 = 100
(x - 4)^2 + (z - 8)^2 = -44
And since there's no possible way to ever get a sum of 2 squares to be equal to a negative number, the answer to this intersection is DNE. This shouldn't be a surprise since the center point is 12 units from this plane and the sphere has a radius of only 10 units.
yz-plane:
(x - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(0 - 4)^2 + (y + 12)^2 + (z - 8)^2 = 100
(-4)^2 + (y + 12)^2 + (z - 8)^2 = 100
16 + (y + 12)^2 + (z - 8)^2 = 100
(y + 12)^2 + (z - 8)^2 = 84</span>
If the endpoints of a diameter are (6,3) and (2,1) the midpoint is the center of the circle so:
(x,y)=((6+2)/2, (3+1)/2)=(4,2)
Now we need to find the radius....the diameter is:
d^2=(6-2)^2+(3-1)^2
d^2=16+4
d^2=20 since d=2r, r=d/2, and r^2=d^2/4 so
r^2=5
The standard form of the circle is (x-h)^2+(y-k)^2=r^2 and we know:
(h,k)=(4,2) from earlier so:
(x-4)^2+(y-2)^2=5
Answer:
-3/4
Step-by-step explanation:
I think you are missing part of the problem. If you had another equation or a value for x you could solve it.
