Consider a circle whose equation is x2 + y2 + 4x - 6y - 36 = 0. Which statements are true? Check all that apply.

A. To begin converting the equation to standard form, subtract 36 from both sides.

B. To complete the square for the x terms, add 4 to both sides.

C. The center of the circle is at (-2, 3).

D. The center of the circle is at (4, - 6).

E. The radius of the circle is 6 units.

F. The radius of the circle is 49 units.

Complete the squaer for both

in

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

center is (h, k) and radius is r

x²+y²+4x-6y-36=0

group x and y terms

(x²+4x) + (y²-6y) - 36=0

take 1/2 of linear cofients and squaer and add negative and positive inside

(x²+4x+4-4) + (y²-6y+9-9) - 36=0

factor perfect square

((x+2) ²-4) + ((y-3) ²-9) - 36=0

distribute

(x+2) ²-4 + (y-3) ²-9-36=0

add4+9+36 to both sides

(x+2) ² + (y-3) ²=49

(x - (-2)) ² + (y-3) ²=7²

center is (-2,3) and radius is 7

answer is C

(not sure for A and B, because they could be, but they are not neccecary)