The
three bisectors of the sides of a triangle intersect in a unique point,
the circumcentre, which is the centre of the circumscribed circle. In order
to prove this statement we must only think that each point of the bisector
of the side *PQ* is equidistant from *P* and *Q*. Analogously
each point of the side *PR* is equidistant from *P* and *R*.
The intersection point *O* of both bisectors is simultaneously equidistant
from *P*, *Q* and *R*. Then
*O* also belongs to the
bisector of the side *QR*. Therefore the three bisectors meet in a
unique point. Because *O* is equally distant from the three vertices,
the circumscribed circle has its centre in *O*. Let us use this condition
to calculate the equation of the circumcentre:

*OP*^{2} = *OQ*^{2} = *OR*^{2}
= *d*^{2}

where *d* is the radius of the circumscribed circle. Using the
position vectors of each point we have:

(*P* - *O*)^{2} = (*Q* - *O*)^{2}
= (*R* - *O*)^{2}

The first equality yields:

*P*^{2} - 2 *P*** ·** *O* + *O*^{2}
= *Q*^{2} - 2
*Q*** · ***O* + *O*^{2}

By simplifying and arranging the terms containing *O* on
the left
hand side:

2 ( *Q* - *P* ) **· ***O* = *Q*^{2}
- *P*^{2
}2 *PQ*** · ***O* = *Q*^{2} - *P*^{2}

From the second equality we find an analogous result:

2 *QR* **· ***O* = *R*^{2} - *Q*^{2}

Now we introduce the geometric product instead of the scalar product in these equations:

*PQ O* + *O PQ* = *Q*^{2} - *P*^{2}

*QR O* + *O QR* = *R*^{2} - *Q*^{2}

By subtraction of the second equation multiplied by *PQ*
on the
right side minus the first equation multiplied by *QR* on the left,
we obtain:

*PQ QR O* - *O PQ QR* = *PQ R*^{2} - *PQ Q*^{2}
- *Q*^{2}*QR* + *P*^{2} *QR*

By using the permutative property on the left hand side and simplifying the right hand side, we have:

*PQ QR O* - *QR PQ O* = *P*^{2} *QR* + *Q*^{2}*RP*
+ *R*^{2} *PQ*

2 ( *PQ* ^ *QR* ) *O* = *P*^{2} *QR*
+ *Q*^{2} *RP* + *R*^{2} *PQ*

Finally, the multiplication by the inverse of the outer product on the left gives:

*O* = ( 2 *PQ* ^ *QR* ) ^{-1} ( *P*^{2}*QR*
+ *Q*^{2}*RP* + *R*^{2} *PQ* )

= - ( *P*^{2}*QR* + *Q*^{2}*RP*
+ *R*^{2} *PQ* ) ( 2 *PQ* ^ *QR* ) ^{-1}

a useful formula for the calculation of the coordinates of the circumcentre. For example, let us calculate the centre of the circle passing through the points:

*P* = ( 2, 2 ) *Q*
= ( 3, 1 ) *R* = (
4, - 2 )

*P*^{2} = 8
*Q*^{2} = 10
*R*^{2} = 20

*QR* = *R* - *Q* = *e*_{1} - 3 *e*_{2}
*RP* = *P* - *R* = - 2 *e*_{1} + 4 *e*_{2}
*PQ* = *Q* - *P* = *e*_{1} - *e*_{2}

*PQ* ^ *QR* = -2 *e*_{12} ( *PQ* ^ *QR*
) ^{-1} = *e*_{12} / 2

*O* = - ( 8 ( *e*_{1} - 3 *e*_{2} ) +
10 ( - 2 *e*_{1} + 4 *e*_{2
}) + 20 ( *e*_{1}
- *e*_{2} ) ) *e*_{12} / 4

= - ( 8 *e*_{1} - 4 *e*_{2}
) *e*_{12} / 4 = - *e*_{1} - 2 *e*_{2}
= ( - 1, - 2 )

In order to deduce the radius of the circle, we take the vector
*OP*:

*OP* = *P* - *O* = *P* + ( *P*^{2 }*QR*
+ *Q*^{2} *RP* + *R*^{2} *PQ* ) ( 2
*PQ*
^ *QR* ) ^{-1}

and extract the inverse of the triangle area as a common factor:

*OP* = ( 2 *P* *PQ* ^ *QR* + *P*^{2 }*QR*
+ *Q*^{2 }*RP* + *R*^{2} *PQ* ) ( 2
*PQ*
^ *QR* )^{ -1}

= [ 2 *P* ( *P* ^ *Q*
+ *Q* ^ *R* + *R* ^
*P* ) + *P*^{2} *QR*
+ *Q*^{2} *RP* + *R*^{2} *PQ* ] ( 2
*PQ*
^ *QR* )^{ -1}

= [ *P* ( *P Q* - *Q P*
+ *Q R* - *R Q* + *R P* - *P R* ) + *P*^{2}
( *R* - *Q* ) + *Q*^{2} ( *P* - *R* )
+ *R*^{2} ( *Q* - *P* ) ]

( 2 *PQ* ^ *QR*
) ^{-1}

The simplification gives:

*OP* = ( *P Q R* - *P R Q* + *P R P* - *P Q P*
+ *Q*^{2} *P* - *Q*^{2} *R* + *R*^{2}*Q*
- *R*^{2}*P* ) ( 2 *PQ* ^ *QR *)^{ -1}

= - (*Q* - *P*) (*R* -
*Q*)
(*P* - *R*) ( 2 *PQ* ^ *QR* )^{ -1} = - *PQ
QR RP* ( 2 *PQ* ^ *QR* )^{ -1}

Analogously:

*OQ* = - *QR RP PQ* ( 2 *PQ* ^ *QR* )^{ -1}
*OR* = - *RP PQ QR* ( 2 *PQ* ^ *QR* )^{ -1}

The radius of the circumscribed circle is the length of any of these vectors:

where we find the law of sines.

The addition of these vectors yields the direction vector of the Euler line of the triangle:

Updated: August 21^{st}, 2010