X 2 +y 2 +4x−2y=−1space, x, start superscript, 2, end superscript, plus, y, start superscript, 2, end superscript, plus, 4, x, m
inus, 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
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.
