General 2^nd degree equation --> standard equation of a circle: Find the center and radius of a circle determined by

We need to rewrite this in the form of the standard equation of a circle. Once we do, we can read the coordinates (h,k) of the center and the value of the radius.

To do this we need to complete the squares by determining what constant terms could be added make the form (x - h)², and a similar form for the y terms. First we move the constant to the right hand side and group the other terms:

(x² - 6x ) + (y² + 8y ) = 11

We see that if we add 9 to the (x² - 6x ) group, we would have the square (x - 3)², and if we add 16 to the (y² +8y ) group we would have the square (y+4)². Of course we need to add a similar total to the right hand side to compensate. Here are the details:

, or

Comparing this with the standard equation of a circle, we see that the center is at (3, -4) and the radius is 6.

Test Problem: Find the center and radius of a circle determined by

a) center: (-4,2), radius = 3

b) center: (4,2), radius = 3

c) center: (-2,1), radius = 11

d) center: (2,1), radius = 4

e) center: (-2,1), radius = 4

f) center: (2,1), radius =