Finding a Circle from Given Points

Often a problem does not hand you the centre and radius directly — instead it describes the circle through the points it passes through. Two classic situations come up again and again.

Circle on a given diameter. Suppose $A(x_1,y_1)$ and $B(x_2,y_2)$ are the two ends of a diameter. For any other point $P(x,y)$ on that circle, the angle $\angle APB$ is a right angle — this is the familiar 'angle in a semicircle' fact. A right angle at $P$ means the segments $PA$ and $PB$ are perpendicular, so the product of their slopes is $-1$, which after clearing denominators gives the neat diameter-form equation

No separate step of finding the centre or radius is needed — this equation already is the circle.

Circle through three non-collinear points. Three points that are not all on one line determine exactly one circle (this is why a wobbly three-legged stool never rocks — three points fix a unique circle, or in 3D a unique plane). To find it algebraically, substitute each of the three given points into the general form $x^2+y^2+2gx+2fy+c=0$. Each substitution gives one linear equation in the unknowns $g$, $f$, $c$; three points give three equations, which you solve simultaneously (or, equivalently, express as a $3\times 3$ determinant condition) to pin down $g$, $f$, $c$ and hence the circle.

Worked Example 1 (diameter form). Find the equation of the circle whose diameter has endpoints $A(1,2)$ and $B(5,-4)$.

Directly apply the diameter formula: $(x-1)(x-5)+(y-2)(y+4)=0$. Expanding, $x^2-6x+5+y^2+2y-8=0$, i.e. $x^2+y^2-6x+2y-3=0$. (As a check, the centre should be the midpoint of $AB$, namely $(3,-1)$; reading off $-g=3,-f=-1$ from the equation confirms $g=-3,f=1$ — matching $2g=-6$ and $2f=2$ above.)

Worked Example 2 (three points). Find the circle through $(0,0)$, $(4,0)$ and $(0,-6)$. …