This is what I wish we had.

It would be easier to fit arcs when you can’t snap to center.

Here is the math if you need it…

Date: 05/25/2000 at 00:14:35

From: Alison Jaworski

Subject: finding the coordinates of the center of a circle

Hi,

Can you help me? If I have the x and y coordinates of 3 points - i.e.

(x1,y1), (x2,y2) and (x3,y3) - how do I find the coordinates of the

center of a circle on whose circumference the points lie?

Thank you.

Date: 05/25/2000 at 10:45:58

From: Doctor Rob

Subject: Re: finding the coordinates of the center of a circle

Thanks for writing to Ask Dr. Math, Alison.

Let (h,k) be the coordinates of the center of the circle, and r its

radius. Then the equation of the circle is:

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

Since the three points all lie on the circle, their coordinates will

satisfy this equation. That gives you three equations:

```
(x1-h)^2 + (y1-k)^2 = r^2
(x2-h)^2 + (y2-k)^2 = r^2
(x3-h)^2 + (y3-k)^2 = r^2
```

in the three unknowns h, k, and r. To solve these, subtract the first

from the other two. That will eliminate r, h^2, and k^2 from the last

two equations, leaving you with two simultaneous linear equations in

the two unknowns h and k. Solve these, and you’ll have the coordinates

(h,k) of the center of the circle. Finally, set:

```
r = sqrt[(x1-h)^2+(y1-k)^2]
```

and you’ll have everything you need to know about the circle.

This can all be done symbolically, of course, but you’ll get some

pretty complicated expressions for h and k. The simplest forms of

these involve determinants, if you know what they are:

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

Example: Suppose a circle passes through the points (4,1), (-3,7), and

(5,-2). Then we know that:

```
(h-4)^2 + (k-1)^2 = r^2
(h+3)^2 + (k-7)^2 = r^2
(h-5)^2 + (k+2)^2 = r^2
```

Subtracting the first from the other two, you get:

```
(h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0
(h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0
h^2+6*h+9 - h^2+8*h-16 + k^2-14*k+49 - k^2+2*k-1 = 0
h^2-10*h+25 - h^2+8*h-16 + k^2+4*k+4 - k^2+2*k-1 = 0
14*h - 12*k + 41 = 0
-2*h + 6*k + 12 = 0
10*h + 65 = 0
30*k + 125 = 0
h = -13/2
k = -25/6
```

Then

```
r = sqrt[(4+13/2)^2 + (1+25/6)^2]
= sqrt[4930]/6
```

Thus the equation of the circle is:

```
(x+13/2)^2 + (y+25/6)^2 = 4930/36
```

- Doctor Rob, The Math Forum

Classroom Resources - National Council of Teachers of Mathematics