If you know 4 points on a circle, how can you get its equation?

Question

Suppose, for example, you have two x-intercepts and two y-intercepts. They're not additive inverses of each other. Is there a way to find the center and radius? Best answer will have a complete explanation.

Answer 1

Three are enough:

The circle's equation is (x - x0)^2 + (y - y0)^2 = r^2

where (x0, y0) is the center and r is the radius.

Form a system of 3 equations by plugging the coordinates of the points into (x, y), solve for x0, y0 and r.

When, you subtract one equations from another you'll find a system of 2 linear equations.

Be (x1, y1), (x2, y2), (x3, y3) the points:

(x1 - x0)^2 + (y1 - y0)^2 = r^2
(x2 - x0)^2 + (y2 - y0)^2 = r^2

(x1)^2 - 2x0*x1 + (x0)^2 + (y1)^2 - 2y0 * y1 + (y0)^2 = r^2
(x2)^2 - 2x0*x2 + (x0)^2 + (y2)^2 - 2y0 * y2 + (y0)^2 = r^2

Subtract the 1st equation from the 2nd

(x2)^2 - (x1)^2 - 2x0(x2 - x1) + (y2)^2 - (y1)^2 -2y0(y2 - y1) = 0

Likewise,
(x3)^2 - (x1)^2 - 2x0(x3 - x1) + (y3)^2 - (y1)^2 -2y0(y3 - y1) = 0

You already have x1, x2, x3, y1, y2, y3 and you need to find x0, y0 by solving a system of linear equations.

After that, finding r^2 will be easy.

Answer 2

You only need three points. The center of the circle is the intersection of the perpendicular bisectors of the points taken pairwise. The radius is distance from the newly found center of the circle to any one of the points on the circle.

Answer 3

Let the Eqn. be

x^2 + y^2 + 2gx + 2fy + c = 0
Put The Value of any three pts. in the above eqn. to get three relations in g,f,c.
Hence solve them to get the eqn.

Also check whether the fourth point satisfies the eqn.

Answer 4

3 points are enough...