Question

Centre of
circle
x^+y^+2gx+2fy+c=0

Answer 1

Step-by-step explanation:

A circle is a locus of a point which moves such that its distance from a fixed point is always constant.

Let the fixed point be (g,f) which is this case is the center of the circle.

Let the variable moving point be (h,k)

From the definition we have:

(h - g)^2 + (k - f)^2 = r (Where r is the fixed I distance. In this case, this is the radius)

=> h^2 - 2gh + g^2 + k^2 - 2kf + f^2 = r

=> h^2 + k^2 - 2gh - 2kf + g^2 + f^2 - r = 0

replace (h,k) with (x,y) and you have the answer

x^2 + y^2 - 2gx - 2fy + g^2 + f^2 - r = 0

Let g^2 + f^2 - r = c

so we have x^2 + y^2 - 2gx - 2fy + c = 0

In some conventions, we take the center of the circle to be (-g, -f). That accounts for the positive sign in the equation.