An excenter is the center of an excircle. An excircle is one of three circles that touches a triangle - one for each side.

Triangle ΔABC has three vertices, **A**, **B**, and **C**, three sides, AB, BC, and CA, and three associated side lengths, c, a, and b, respectively. A scalar quality called the semiperimeter is an *extremely* useful quantity that shows up repeatedly in the analytical geometry of circles An example is Heron's formula for the equation of the area of the triangle, K, which is sqrt(s*(s-a)*(s-b)*(s-c)). Another example is that the radius of the incircle, r, is simply equal to K/s. No square roots. No integrals. No complicated engineering functions.

It turns out that if you extend the triangle's sides by the right amount, then an excircle will fit to three points, where two of the points are gotten from the triangle side's extension, while the third point is the contact point on the triangle itself. And magically, it turns out that the contact point and the two lengths of the extensions of the sides of the triangle are all related to the quantities (s-a), (s-b), and (s-c), the same terms you've seen in Heron's formula above.

Heron (ca 10-70 AD) was a mathematician who lived in Alexandria, Egypt, which was then under Roman control. Near as I can tell, he looked like the actor Gerard Butler, but probably did not date women like Jenifer Aniston. Hey, I'm just speculating. He was a quite prolific inventor and a pretty good mathematician as well. We don't know a lot about him; some of his manuscripts were preserved by Arab scholars. Not only was he responsible for this brilliant little equation about the area of a triangle based on just the side lengths, but he also developed an iterative algorithm that could compute square roots. He also developed a rotating device that shot out jets of steam as water came to a boil, a water pump, and a syringe for pushing water or air through a pipe. The dude was fascinated by fluids.

Let's find the centers of the three excircles, Γ_{A}, Γ_{B}, and Γ_{C}. The first way is with compass and straight-edge construction, and the second way will be analytically.

**Compass and Straightedge Construction**: Do you remember how you constructed the center of the incircle? The incenter **O** was the intersection of the angle bisectors of the interior angles.

In like manner, the excenter **O**_{A} is the intersection of the angles formed by two exterior angles. If you extend sides a and b, the excircle Γ_{A} touches side extensions a and b, and also touches the side c. The incenter is the intersection of two angle bisectors of exterior angles. One is the exterior angle at vertex **A**, which is the supplementary angle of ∠A (its angle is 180&Deg; - ∠A). The other is the exterior angle of **C**. When you create the angle bisectors of these exterior angles at **A** and **C**, the origin of Γ_{A} lies at the intersection.

This was known to the ancients. Euclid demonstrates it in The Elements in Book IV, Prop. 4

**Analytic Method for Finding Coordinates of Excenters**: If you know the triangle's vertex coordinates, you can find the centers of the excircles using a function that fits a circle to three points. Fortunately, there is one. It's the `CIRCLE` function, which I've described in A circle is defined by three points.

First, I'll give you the geometrically visual of what's going on. Then you can find the formulae for calculating the excenters' coordinates. Construct triangle ΔABC and measure (or calculate) the side lengths. Now you have a,b,c. Compute the semiperimeter s. Calculate the following quantities: (s-a), (s-b) and (s-c). Now let's go back to the triangle you've drawn and extend each side length by an appropriate length:

- Extend side a
- Along the ray BC, subtract a length of distance (s-a) and call that point
**B**_{C}
- By "subtract" I mean that since the ray is pointing from
**B** toward **C**, point **B**_{C} is pointing away from **B**.
- The distance between
**B** and **B**_{C} is s-a.
- This point will be used to create excircle Γ
_{C}

- Along the ray BC, add a length of distance (s-a) and call that point
**C**_{B}
- This point will be used to create excircle Γ
_{B}

- On this line segment there are four points:
**B**_{C}, **B**, **C**, and **C**_{B}
- The length of line segment
**B****C** is a
- The length of the extended line segment
**B**_{C}**C**_{B} is b+c

- Extend side b
- Along the ray CA, subtract a length of distance (s-b) and call that point
**C**_{A}.
- Since the ray is pointing toward
**A**, point **C**_{A} is pointing away from **B**, and is further from **A** than **C**
- The distance between
**C** and **C**_{A} is s-b.
- This point will be used to create excircle Γ
_{A}

- Along the ray CA, add a length of distance (s-b) and call that point
**A**_{C}
- Since the ray is pointing toward
**A**, point **A**_{C} is pointing in the same direction as **A**, and is further from **C** than **A**
- The distance between
**C** and **A**_{C} is b + (s-b) = s.
- This point will be used to create excircle Γ
_{C}

- On this line segment there are four points:
**C**_{A}, **C**, **A**, and **A**_{C}
- The length of line segment
**C****A** is b
- The length of the extended line segment
**C**_{A}**A**_{C} is (s-b) + b + (s-b) = a+c

- Extend side c
- Along the ray AB, add a length of distance (s-c) and call that point
**A**_{B}
- This point will be used to create excircle Γ
_{B}

- Along the ray AB, add a length of distance (s-c) and call that point
**B**_{A}
- This point will be used to create excircle Γ
_{A}

- You're done with this step

At this point you have six of the nine points you need for the three excircles. You now need to draw/calculate the touchpoints on ΔABC where the excircles touch the triangle. You do this by starting at a vertex and finding the distance from that vertex where the exterior touch point is located. By a miracle, it's always one of these same distances we've used before: (s-a), (s-b), or (s-c). I supposed I oughtn't to say "by a miracle." I ought to pretend that I knew this all along. But that would be contrary to the surprise I remember feeling when I first found out this fact. It surprised the hell out of me that these touch points could be computed using the same semiperimeter distances that appear in Heron's formula.

Let's begin at vertex **B** and work our way around the triangle counterclockwise.

- Find exterior touch point
**T**_{A}'.
- Begin at vertex
**B**
*Reminder: This is the point at which the excircle Γ*_{A} touches side BC.
- Along the ray BC, travel in the direction of vertex
**C** a length (s-c) and label that point **T**_{A}'

- Find exterior touch point
**T**_{B}'.
- Move to vertex
**C**
*Reminder: The excircle Γ*_{B} touches side CA at point **T**_{B}'
- Along the ray CA, travel in the direction of vertex
**A** a length (s-a) and label that point **T**_{B}'

- Find exterior touch point
**T**_{C}'.
- Move to vertex
**A**
*Reminder: The excircle Γ*_{C} touches side AB at **T**_{C}'
- Along the ray AB, travel in the direction of vertex B a length (s-b) and label that point
**T**_{C}'

- You're done with this step

It's not so obvious unless you've drawn this a number of times, but the distances (s-a), (s-b), and (s-c) are all less than any one side length. That's why the list of instructions above works without having to check if you've passed the second vertex or not.

Summary: **Equations for coordinates of the the points of tangency and the extension points**:

For excircle Γ_{A}
------------------------------------
**T**_{A}' = **B** + ((s-c)/a)* (**C** - **B**) (1a)
T_A'x = Bx + ((s-c)/a) * (Cx-Bx)
T_A'y = By + ((s-c)/a) * (Cy-By)
**B**_{A} = **A** + ((1+(s-c))/c)* (**B** - **A**) (1b)
B_Ax = Ax + ((1+(s-c))/c) * (Bx-Ax)
B_Ay = Ay + ((1+(s-c))/c) * (By-By)
**C**_{A} = **C** - ((s-b))/b)* (**A** - **C**) (1c)
C_Ax = Cx - ((s-b)/b) * (Ax-Cx)
C_Ay = Cy - ((s-b)/b) * (Ay-Cy)
For excircle Γ_{B}
------------------------------------
**T**_{B}' = **C** + ((s-a)/b)* (**A** - **C**) (2a)
T_B'x = Cx + ((s-a)/b) * (Ax-Cx)
T_B'y = Cy + ((s-a)/b) * (Ay-Cy)
**C**_{B} = **B** + ((1+(s-a))/a)* (**C** - **B**) (2b)
C_Bx = Bx + ((1+(s-a))/a) * (Cx-Bx)
C_By = By + ((1+(s-a))/a) * (Cy-By)
**A**_{B} = **A** - ((s-c))/c)* (**B** - **A**) (2c)
A_Bx = Ax - ((s-c)/c) * (Bx-Ax)
A_By = Ay - ((s-c)/c) * (By-Ay)
For excircle Γ_{C}
------------------------------------
**T**_{C}' = **A** + ((s-b)/c)* (**B** - **A**) (3a)
T_C'x = Ax + ((s-b)/c) * (Bx-Ax)
T_C'y = Ay + ((s-b)/c) * (By-Ay)
**A**_{C} = **C** + ((1+(s-b))/b)* (**A** - **C**) (3b)
A_Cx = Cx + ((1+(s-b))/b) * (Ax-Cx)
A_Cy = Cy + ((1+(s-b))/b) * (Ay-Cy)
**B**_{C} = **B** - ((s-a))/a)* (**C** - **B**) (3c)
B_Cx = Bx - ((s-a)/a) * (Cx-Bx)
B_Cy = By - ((s-a)/a) * (Cy-By)

At this point you have three points for each excircle. Now you can find the centers and radii of each of the excircles, using the CIRCLE function defined in A circle is defined by three points to fit a circle through three points.

- Excircle A fits through (
**T**_{A}', **B**_{A}, **C**_{A})
- The excenter for Γ
_{A} and radius r_{A} are:
`(O_Ax, O_Ay, r_A) = circle(T_Ax', T_Ay', B_Ax, B_Ay, C_Ax, C_Ay)`

- Excircle B fits through (
**T**_{B}', **C**_{B}, **A**_{B})
- The excenter for Γ
_{B} and radius r_{B} are:
`(O_Bx, O_By, r_B) = circle(T_Bx', T_By', C_Bx, C_By, A_Bx, A_By)`

- Excircle C fits through (
**T**_{C}', **A**_{C}, **B**_{C})
- The excenter for Γ
_{C} and radius r_{C} are:
`(O_Cx, O_Cy, r_C) = circle(T_Cx', T_Cy', A_Cx, A_Cy, B_Cx, B_Cy)`

**Example 1**: Find a triangle's three excenters and radii given the coordinates of its vertices.

**GIVEN**: The vertices of ΔABC:
A = (Ax, Ay) = ( 2, 5)
B = (Bx, By) = ( 6, 3)
C = (Cx, Cy) = ( 4, 6)
**FIND**: the locations and radii of the three excircles Γ_{A}, Γ_{B}, and Γ_{C}
**SOLUTION** The reference triangle's basic information (side lengths, radius of incircle and circumcircle, etc:
a = 3.6
b = 2.2
c = 4.5
s = 5.2
K = 4
r = 0.8
R = 2.3
O = (Ox, Oy) = ( 3.7, 5.0)
O' = (O'x, O'y) = ( 3.9, 3.8)
The excenters and radii are:
O_A = (O_Ax, O_Ay) = ( 7.8, 5.0)
O_B = (O_Bx, O_By) = ( 2.0, 6.5)
O_C = (O_Cx, O_Cy) = ( 2.0,-1.5)
r_A = 2.6
r_B = 1.4
r_C = 5.8
**CHECK**: K = sqrt(r⋅r_A⋅r_B⋅r_C) (4)
K = 4
sqrt(r⋅r_A⋅r_B⋅r_C) = sqrt(2.3 ⋅ 2.6 ⋅ 1.4 ⋅ 5.8) = 4
*Check!*

**Barycentric Coordinates**: The excenters **O**_{A}, **O**_{b}, and **O**_{C} have barycentric coordinates:

**O**_{A} ↠ -a : b : c (5a)
**O**_{B} ↠ a : -b : c (5b)
**O**_{C} ↠ a : b : -c (5c)

Excenters are found by converting barycentric coordinates to Cartesian coordinates.
See barycentric coordinates for details.

**O**_{A} = (-a/(-a+b+c))***A** + (b/(-a+b+c))***B** + (c/(-a+b+c))***C** (6a)
**O**_{B} = (a/(a-b+c))***A** + (-b/(a-b+c))***B** + (c/(a-b+c))***C** (6b)
**O**_{C} = (a/(a+b-c))***A** + (b/(a+b-c))***B** + (-c/(a+b-c))***C** (6c)

**Example 2**: Continuing from the previous example, the excenters are found to be:

A = ( 2, 5)
B = ( 6, 3)
C = ( 4, 6)
a = 3.6
b = 2.2
c = 4.5
a b c
O_Ax = - ------*Ax + ------*Bx + ------*Cx
-a+b+c -a+b+c -a+b+c
3.6 2.2 4.5
= - ------------*2 + ---*6 + ---*4
-3.6+2.2+4.5 3.1 3.1
= -1.2*2 + 0.7*6 + 1.4*4
= 7.8
a b c
O_Ay = - ------*Ay + ------*By + ------*Cy
-a+b+c -a+b+c -a+b+c
= -1.2*5 + 0.7*3 + 1.4*6
= 5.0
O_Bx = 2.0
O_By = 6.5
O_Cx = 2.0
O_Cy = -1.5