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### Find Circle Center and Radius

```
Date: 09/21/2001 at 08:51:17
From: Alex
Subject: Geometry

If you are given three sets of (x,y) coordinates that lie on the
circumference of a circle, how do you figure out the center and radius
of the circle?
```

```
Date: 09/21/2001 at 13:43:40
From: Doctor Rob
Subject: Re: Geometry

Thanks for writing to Ask Dr. Math, Alex.

You can start with the three-point form of the equation of the circle,
which can be found on the following page from our Frequently Asked
Questions (FAQ):

http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html#twocircles

| x^2+y^2  x  y  1|
|x1^2+y1^2 x1 y1 1|
|x2^2+y2^2 x2 y2 1| = 0.
|x3^2+y3^2 x3 y3 1|

Here (x1,y1), (x2,y2), and (x3,y3) are the three given points. Expand
this 4-by-4 determinant, complete the squares on x and y, and
transform it into the center-radius 1form

(x-h)^2 + (y-k)^2 = r^2.

Then the center is (h,k) and the radius is r.

Explicit expressions for h, k, and r^2 are as follows:

|x1^2+y1^2 y1 1|
|x2^2+y2^2 y2 1|
|x3^2+y3^2 y3 1|
h = ----------------,
|x1 y1 1|
2*|x2 y2 1|
|x3 y3 1|

|x1 x1^2+y1^2 1|
|x2 x2^2+y2^2 1|
|x3 x3^2+y3^2 1|
k = ----------------.
|x1 y1 1|
2*|x2 y2 1|
|x3 y3 1|

|x1 y1 x1^2+y1^2|
|x2 y2 x2^2+y2^2|
|x3 y3 x3^2+y3^2|
r^2 = h^2 + k^2 + -----------------.
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|

In expanded form, these are:

(x1^2+y1^2)(y2-y3) + (x2^2+y2^2)(y3-y1) + (x3^2+y3^2)(y1-y2)
h = ------------------------------------------------------------
2(x1y2 - x2y1 - x1y3 + x3y1 + x2y3 - x3y2)

(x1^2+y1^2)(x3-x2) + (x2^2+y2^2)(x1-x3) + (x3^2+y3^2)(x2-x1)
k = ------------------------------------------------------------
2(x1y2 - x2y1 - x1y3 + x3y1 + x2y3 - x3y2)

(x1^2+y1^2)(x2y3-x3y2) +
(x2^2+y2^2)(x3y1-x1y3) +
(x3^2+y3^2)(x1y2-x2y1)
r^2 = h^2 + k^2 + ---------------------------------------
x1y2 - x2y1 - x1y3 + x3y1 + x2y3 - x3y2

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/17/2002 at 10:31:05
From: Michael Greene
Subject: Equation for a circle given (x,y) of 3 points

I have posed this question to geometry and Algebra 2 classes for the
past 3-4 years, and the geometric solution may be simpler for many
students who recall the geometry and may not be as comfortable with
systems of equations:

Consider A(-2,1) B(2,3) C(0,-5)

a) The center of a circle may be found by the intersection of the
perpendicular bisectors of 2 chords.

b) Find the slopes and midpoints of 2 chords:
AB: (0,2) m = 1/2
BC: (1,-1) m = 4

c) Find the slopes of the perpendicular lines.

d) Find the equations of the perpendicular bisectors:
Through AB, y = -2x+2
Through BC, y = (-1/4)x  -3/4

e) Find the intersection of those lines
-2x +2 = (-1/4)x -3/4
-8x +8 = -x -3
-7x = -11
x = 11/7 and y = -8/7

This isn't so much a question as an alternate solution, but I wanted
to give something back to the Math Forum after referring some of my
students to the archives and finding new ideas for our classes.

Thanks,
Mike Greene
```
Associated Topics:
High School Conic Sections/Circles
High School Coordinate Plane Geometry
High School Geometry

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