Given the equation of a circle"x^2 + y^2 - 4x +6y -36 = 0We will use completing the square method to rewrite the equation into the form:(x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center and r is the radius.Let us rewrite the terms.==> x^2 - 4x + y^2 + 6y = 36Now we will complete the square for both x^2 ands y^2.We will add [( coefficient of x)/2]^2 and [(coefficients of y)/2]^2 to both sides.Then we will add :(4/2)^2 = 2^2 = 4(6/2)^2 = 3^2 = 9Then we will add 4 and 9 to both sides.==> x^2 - 4x +4 + y^2 + 6y + 9 = 36 + 4 + 9==> (x-2)^2 + (y+3)^2 = 49==> (x-2)^2 + (y+3)^2 = 7^2Now we will compare the equation withe the standard form of a circle.Then we conclude that:The center of the circle is: ( 2, -3) and the radius is 7.