Circle through three points

Find the circle through the three points $(0, 0)$ , $(4, 0)$ and $(0, 2)$ . Through any three points not on a single line, exactly one circle passes¹.

Using the general form

Write the circle as $x^2 + y^2 + \ell x + m y + n = 0$ and substitute each point.

Point	Equation	Result
$(0, 0)$	$n = 0$	$n = 0$
$(4, 0)$	$16 + 4\ell = 0$	$\ell = -4$
$(0, 2)$	$4 + 2m = 0$	$m = -2$

Three unknowns, three conditions: exactly enough to pin the circle down. This gives $x^2 + y^2 - 4x - 2y = 0$ , and completing the square yields the standard form.

(x - 2)^2 + (y - 1)^2 = 5

The centre is $(2, 1)$ and the radius is $\sqrt{5}$ .

Using perpendicular bisectors

A completely different route reaches the same place. The centre must be equidistant from all three points, and the set of points equidistant from two given points is the perpendicular bisector of the segment joining them.

Pair of points	Perpendicular bisector
$(0, 0)$ and $(4, 0)$	$x = 2$
$(0, 0)$ and $(0, 2)$	$y = 1$

They cross at the centre we already found, and the radius is the distance from there to $(0, 0)$ , namely $\sqrt{5}$ .

Why the circle is unique

The centre is fixed as the intersection of two perpendicular bisectors, and once the centre is fixed so is the radius. If the three points do lie on one line, those bisectors are parallel and never meet, and no circle passes through all three. The third bisector always passes through the same intersection, so the condition of being equidistant from all three points is consistent.

The circumscribed circle

The circle is the circumscribed circle of the triangle with those vertices, and its centre is the circumcentre. The familiar fact that a triangle's circumcentre is the meeting point of the perpendicular bisectors of its sides has just been confirmed in coordinates.

The large dots mark the three points and the centre.

Circumcircle, Wikipedia