Find the center and radius of this circle

x^2 + y^2 + 22x - 28y + 196 = 0

I know that the formula for a circle is (x-h)^2 + (y-k)^2 = r^2 but i just don't know how to put it into that formula. Thanks!

You have to regroup and complete the square to factor.

X^2+ 22x + y^2 - 28y= -196

For the x: half of 22 is 11, then 11^2= 121
For y: half of 28 is 14, 14^2 is 196
So add 121 to the x grouping, and 196 to the y. Add also to the right side.

X^2+ 22x + 121+ y^2 - 28y + 196= -196 + 121 + 196. Now factor and simplify

(x+ 11)^2 + (y-14)^2 = 121

C(-11,14) r = 11

Hoping this helps!

you have to complete the square(s)...

(x^2 + 22x + ___) + (y^2 - 28y + ___) = - 196

(x^2 + 22x + 121) + (y^2 - 28y + 196) = - 196 + 196 + 121

(x + 11)^2 + (y - 14)^2 = 121

can you find the center and radius ? I hope so...read your textbook...practice...practice a lot !

[center: (- 11, 14), r = 11]

(-11,14) CENTER
R=11