How do you find center and radius for x^2+y^2-12x+8y+51=0

We know the standard form of equation for the circle is:
(x – a)2 + (y – b) = r2 :where the point (a,b) is the centre of the center and r is the radius of the circle.

So lets try to rewrite in the standard form. we get:
x^2+y^2-12x+8y+51=0
x^2-12x + 6^2 -6^2 +y^2+8y +4^2-4^2 + 51 = 0
(x-6)^2 + (y+4)^2 -6^2 -4^2+51 = 0
(x-6)^2 + (Y+4)^2 -1 = 0
(x-6)^2 + (Y+4)^2 = 1 = 1^2
now, if we compare with the standard form, we get:

x = 6 and y = -4 and r = 1
ans: center point is (6,-4) and radius is 1

complete the square for x and y:
(x² – 12x + ....) + (y² + 8y + ....) = -51
(x² – 12x + 36) + (y² + 8y + 16) = -51 + 36 + 16
(x – 6)² + (y + 4)² = 1

so center is (6, -4) and radius is 1

x^2 + y^2 - 12x + 8y + 51 = 0
x^2 - 12x + 36 + y^2 + 8y + 16 + 51 = 36 + 16
(x - 6)^2 + (y + 4)^2 + 51 = 52
(x - 6)^2 + (y + 4)^2 = 1
radius is 1
center is (6, -4)

x^2+y^2-12x+8y+51=0

x^2+y^2-12x+8y+36+15=0

x^2-12x+36+y^2+8y+16-1=0

(x-6)^2+(y+4)^2=1

ie
center(x,y)=(6,-4) and radius is 1