Help with points of intersection of the circle?

Question

What is the process of making this equation into a circle (equation of two halves)?
(x^2)+(y^2)-5x+4y-13=0.

Thanks for taking your time helping me out :)

Answer 1

x^2-5x +y^2+4y-13=0
(x-5/2)^2+(y+2)^2-25/4-13-4=0
x-5/2)^2+(y+2)^2=23+1/4=93/4
center (5/2,-2)
radius=sqrt(93)/2

Answer 2

x^ -5x + y^2 + 4y = 13
x^2 - 5x + (5/2)^2 + Y^2 + 4y + 2^2 = 13 - (5/2)^2 - 2^2
(x - 5/2)^2 + (y + 2)^2 = 13 - 25/4 - 4
(x - 5/2)^2 + (y + 2)^2 = 9 - 25/4
(x - 5/2)^2 + (y + 2)^2 = (36 - 25 ) / 4 = 11 / 4

center of circle ( 5/2, -2 )
radius = sqr(11) / 2

Answer 3

Find coordinates of radius of circle first
Eg : let it be (a,b)
Find lenth of radius using length segment formula
Let any point on curve of circle be (x,y)
(x-a)^2+(y-b)^2= (radius)^2
Simpify and rearrange equation in the form of (x^2)+(y^2)-5x+4y-13=0 to get equation