X2 +y 2 +4x−2y=−1space, x, start superscript, 2, end superscript, plus, y, start superscript, 2, end superscript, plus, 4, x, minus, 2, y, equals, minus, 1 the equation of a circle in the xyxyx, y-plane is shown above. what is the radius of the circle? choose 1 (choice a. a 2

The radius is 2.  

We will rewrite this equation in center-radius form,
(x-h)²+(y-k²)=r²

where (h, k) is the center of the circle and r is the radius.

In order to do this, we will need to complete the square.  We will first rewrite this with the x terms grouped together and the y terms grouped together:
x²+4x+y²-2y = -1

To complete the square for the x terms, we divide the coefficient b (as in bx) by 2:

4/2 = 2

Now we square this:
2²=4

We will add this to both sides of the equation (in order to maintain balance, we must add it to both sides):
x²+4x+4+y²-2y=-1+4
x²+4x+4+y²-2y=3

We have completed the square for the x terms.  When we divided b by 2, that gave us the number we need, 2:
(x+2)²+y²-2y=3

Now we will do the same thing for the y terms.  The b for our y terms is -2:

-2/2 = -1
(-1)²=1

So we will add 1 to both sides, and use -1 in the finalized form:
(x+2)²+y²-2y+1=3+1
(x+2)²+(y-1)²=4

We can see that the center would be located at (-2, 1) and the radius is √4=2.