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```
Date: 09/05/97 at 15:02:19
From: Ron Culpepper

Given three coplanar points, what is the formula for the radius?
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```
From: Dr. Math [SMTP:dr.math@mathforum.org]
Sent: Saturday, September 06, 1997 6:56 PM

Hi Ron,

Do you mean the circle through the three points?

If the points P1, P2, and P3 are in the (x,y)-plane, you could write
the equation of the perpendicular bisector of the line joining P1 and
P2 and do the same for the line joining P2 and P3. The intersection of
the two perpendicular bisectors is the center of the circle. Then you

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 09/09/97 at 15:43:57
From: Ron Culpepper

Yes the circle through three points.

Given p1,p2,p3 and
B1 = (p1-p2)\2
B2 = (p1-p3)\2
B3 = (p2-p3)\2

what is the equation for the perpendicular bisector intersection or
where can I find it?
```

```
Date: 09/15/97 at 12:28:40
From: Doctor Rob

The line connecting P1 and P2 is given by the two-point form of the
equation of a line:

(y-y1)*(x2-x1) = (y2-y1)*(x-x1)

and similarly for P1 and P3, and for P2 and P3. The midpoint of P1P2
is ([x1+x2]/2,[y1+y2]/2). The perpendicular bisector must pass through
this point and have slope the negative reciprocal of the slope of the
line P1P2, that is, slope -(x2-x1)/(y2-y1).

By changing the subscripts, the same holds true for the lines P1P3 and
P2P3. This will allow you to write down the equations for the three
perpendicular bisectors using the point-slope form of the equation of
a line: if the point is (x0,y0) and the slope is m, then the equation
is y = m*(x-x0)+y0.  (You may have to deal with vertical lines, whose
perpendicular bisectors are horizontal and so have slope zero, and
horizontal lines, whose perpendicular bisectors are vertical and so
have the form x = constant.)

Now pick any two of these equations, and solve them simultaneously
for x and y. That will give you the center of the circle. Find the
distance of that point from any of the three given points to get the

Having said that, there is a formula for the radius you seek in terms
of the lengths of the three sides. Let their lengths be a, b, and c.
Let s = (a+b+c)/2, the semi-perimeter.  Then the radius is

R = a*b*c/(4*Sqrt[s*(s-a)*(s-b)*(s-c)]).

You may recognize the square root in the denominator as the area
of the triangle, according to Hero's Formula.  You have to compute
a, b, and c by using the distance formula.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Coordinate Plane Geometry
High School Equations, Graphs, Translations
High School Geometry

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