Question

(1 point) Answer the following questions about the sphere whose equation is given by x2+y2+z2−4x+4y=73. x2+y2+z2−4x+4y=73. 1. Find the radius of the sphere. Radius: r=r= 2. Find the center of the sphere. Write the center as a point (a,b,c)(a,b,c) where aa, bb, and cc are numbers.

Answer 1

Answer:

R=9.

C=(2,-2,0).

Explanation:

Lets try to find how $x^2+y^2+z^2-4x+4y=73$ can be written in the form of $(x-a)^2+(x-b)^2+(x-c)^2=R^2$, since then its radius R and center of coordinates C=(a,b,c) would be easily recognizable.

Since we have the terms -4x and +4y, they seem to have come from the terms $(x-2)^2$ and $(y+2)^2$, and we would need to take care of the extra +4 by substracting then. More explicitly:

$(x-2)^2-4=x^2-4x+4-4=x^2-4x$, as we had originally with the x terms

$(y+2)^2-4=y^2+4y+4-4=y^2+4y$, as we had originally with the y terms

Which means:

$x^2-4x+y^2+4y+z^2=(x-2)^2-4+(y+2)^2-4+z^2$

Or, putting all together:

$(x-2)^2-4+(y+2)^2-4+z^2=73$

$(x-2)^2+(y+2)^2+z^2=9^2$

So our radius is 9 and the center is (2,-2,0).