X 2 + 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 answer: (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.