Given the equation of a circle"

x^2 + y^2 - 4x +6y -36 = 0

We 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 = 36

Now 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 = 9

Then 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 = 7^2

Now 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.

We can write x^2 + y^2 - 4x + 6y - 36 = 0 in the form ( x- a)^2 + (y-b)^2 = r^2, where the center of the circle is (a,b) and the radius is r.

x^2 + y^2 - 4x + 6y - 36 = 0

=> x^2 - 4x + 4 + y^2 + 6y + 9 = 36 + 4 + 9

=> (x - 2)^2 + (y + 3)^2 = 49

=> (x - 2)^2 + (y + 3)^2 = 7^2

Here a = 2, b = -3 and r = 7.

Therefore the center is ( 2, -3) and the radius is 7.