# 3 point circle/arc creation

**URL:** <https://discourse.shapr3d.com/t/3-point-circle-arc-creation/23961>\
**Category:** Feature requests\
**Created:** [February 5, 2023, 6:50pm UTC](https://discourse.shapr3d.com/t/3-point-circle-arc-creation/23961 "2023-02-05T18:50:45Z")\
**Posts on this page:** 1\
**Page:** 1

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**Author:** ![Coffee](https://sea2.discourse-cdn.com/flex002/user_avatar/discourse.shapr3d.com/coffee/32/17451_2.png) [@Coffee](https://discourse.shapr3d.com/u/Coffee)\
**Post date:** [February 5, 2023, 6:50pm UTC](https://discourse.shapr3d.com/t/3-point-circle-arc-creation/23961/1 "2023-02-05T18:50:45Z")

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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](http://mathforum.org/dr.math/)

```auto

```