Center and radius of a circle?

The general equation of the circle is:

x^2+y^2+2gx+2fy+c=0

Then its center is (-g,-f) and radius is root(g^2+f^2-c).

The given equation is:

x^2+y^2+4x+6y-3=0

So, comparing with standard equation,

2g=4, 2f=6

g=2, f=3

So, center = (-2, -3)

Radius = root(4+9+3) = 4

x^2+4x+4+y^2+6y+9=3+4+9

(x+2)^2+(y+3)^2=16

center is (-2,-3) and

radius is 4(

(x-a)² + (y-b)² = r² is a circle radius r centered at (a, b).

x²+y²+4x+6y-3

= x²+4x+4 + y²+6y+9 - (4+9+3) so

(x-(-2))² + (y-(-3))² = 4²

this is a circle of radius = 4, centred on (-2, -3)

(x + 2)^2 + (y + 3)^2 = x^2 + y^2 + 4*x + 6*y + 4 + 9

(Choose the constants to give 4*x and 6*y)

= x^2 + y^2 + 4*x + 6*y + 13

= (x^2+y^2+4x+6y-3) + 16 = 0 + 16 = 16

Thus (x + 2)^2 + (y + 3)^2 = 16 = r^2

Thus centre is (-2,-3) and radius is sqrt(16) = 4.<<<

4 units is the radius

center (-2, -3)

radius 4

(x+2)^2 + (x+3 )^2 = 16