Find the center and radius for the circle, given the equation: x^2+16x+y^2-6y=-12

In form

(x-h) ^2 + (y-k) ^2=r^2

center is (h, k) and radius is r

complete the square for both

group x and y seprately

(x^2+16x) + (y^2-6y) = - 12

take 1/2 of each linear coefient and square it then add positive and negative inside parentahsees

16/2=8, 8^2=64

-6/2=-3, (-3) ^2=9

(x^2+16x+64-64) + (y^2-6y+9-9) = - 12

complete the square

((x+4) ^2-64) + ((y-3) ^2-9) = - 12

expand

(x+4) ^2-64 + (y-3) ^2-9=-12

add 64+9 to both sides

(x+4) ^2 + (y-3) ^2=61

or

(x - (-4)) ^2 + (y-3) ^2 = (√61) ^2

center is (-4,3) and radius is √61