Finding the Equation of a Circle from Given Conditions

Depending on what is given, the equation of a circle can be built in different ways, all of which end up in the same general form $x^2+y^2+2gx+2fy+c=0$.

Centre and radius given directly. Substitute into $(x-h_0)^2+(y-k_0)^2=r^2$ and expand.

Endpoints of a diameter given, say $(x_1,y_1)$ and $(x_2,y_2)$. Since the angle in a semicircle is a right angle, any point $P(x,y)$ on the circle satisfies $PA \perp PB$ where $A,B$ are the diameter's endpoints — algebraically, the product of the two slopes from $P$ to $A$ and from $P$ to $B$ is $-1$, which simplifies neatly to:

Three points given. Substitute all three points into the general equation $x^2+y^2+2gx+2fy+c=0$ to get three simultaneous linear equations in the unknowns $g$, $f$, $c$, then solve them. This works because three non-collinear points determine a unique circle. …