Does anyone know how to find the circumcentre of a triangle using vectors?

its just for any triangle with sides are a, b, c and vertices (x1,y1), (x2,y2) and (x3,y3). are there any theorems i could look up?

Does anyone know how to find the circumcentre of a triangle using vectors?
I don't know the theorem, but the circumcentre is :


( (x1+x2+x3)/3 , (y1+y2+y3)/3 ). If you treat the vertices as a vector, then the vector for the circumcentre is


1/3* ( (x1,y1) + (x2,y2) +(x3,y3) )
Reply:I am not from an English speaking country, so it is sometimes difficult to understand the meaning of some terms.





As I understand, a circumcenter is a center of a circle which comes through the three points A(x1, y1), B(x2,y2), C(x3,y3).





The previous answer gave the center of mass of a triangle with these points as apexes, (x1 + x2 + x3)/3 and (y1 + y2 + y3)/3.





The center of a circle wich comes through the points A, B, C is equally remote from the three points. Geometrically, it is the intersection point of lines that are perpendicular and come across the middle points of sections AB, BC and/or AC.





To find it, we can write and solve two equations for the lines that come across the middles of triangle's sides and perpendicular to the sides.





So, for the AB side, the middle point is {(x1+x2)/2, (y1+y2)/2}. The coefficient k for the line y=kx + a we can find from the condition that for two perpendicular to each other lines


Y = KX + A and y = kx + a, K*k = -1, and we know the coefficient for the AB side of the triangle as (y2-y1)/(x2-x1), hence, the line is


y = - (x2-x1)/(y2-y1)x + a, and to find a we have to take


x = (x1+x2)/2, and to obtain y = (y1+y2)/2:





(y1+y2)/2 = - [(x2-x1)/(y2-y1)](x1+x2)/2 + a


a = (y1+y2)/2 + (x2-x1)(x2+x1)/[2(y2-y1)].





So, the first line is:


y(AB) = - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1))





The second line we shall write for the perpendicular to the BC side of the triangular that comes across the middle of the BC side. Making the analogous calculations, we shall obtain


y(BC)= - (x3-x2)/(y3-y2) x + (y2+y3)/2 + (x3-x2)(x3+x2)/(2(y3-y2))





And the intersection of these two lines will give us the centre of the circle. So, we only have to find the x and y common to these two equations. We have to find x, at which y(AB) = y(BC)


- (x3-x2)/(y3-y2) x + (y2+y3)/2 + (x3-x2)(x3+x2)/(2(y3-y2)) =


= - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1))


x[(x2-x1)/(y2-y1) - (x3-x2)/(y3-y2)] = [(y1+y2) - (y3+y2)]/2 +


(x2-x1)(x2+x1)/(2(y2-y1)) - (x3-x2)(x3+x2)/(2(y3-y2)), and





x[y3(x2-x1) +y1(x3-x2) +y2(x1-x3)]/[(y2-y1)(y3-y2)] = (y1-y3)/2 +


+ (x2-x1)(x2+x1)/2(y2-y1) - (x3-x2)(x3+x2)/2(y3-y2),


or


x = (1/2)[(y2-y1)(y3-y2)(y1-y3) + (x2-x1)(x2+x1)(y3-y2) - (x3-x2)(x3+x2)(y2-y1)]/[y3(x2-x1) + y1(x3-x2) + y2(x1-x3)] =


= X = (1/2)[(y2-y1)(y3-y2)(y1-y3) + y3(x2^2-x1^2) +y1(x3^2-x2^2) +y2(x1^2-x3^2)]/[y3(x2-x1) + y1(x3-x2) + y2(x1-x3)]





This is the x value for the centre of the circle. And we can find the y value for the centre of the circle. For example, from y(AB) = - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1)), we shall input the value of x and obtain:


Y= - 0.5 [(x2-x1)(x1-x3)(x3-x2) - y3^2(x2-x1) - y1^2(x3-x2) - y2^2(x1 - x3)]/[y3(x2 - x1) + y1(x3 - x2) + y2(x1 - x3)] =


= (1/2)[(x2-x1)(x1-x3)(x3-x2) - y3^2(x2-x1) - y1^2(x3-x2) - y2^2(x1-x3)]/[x1(y3-y2) +x2(y1-y3) + x3(y2-y1)] =


= Y = (1/2)[(x2-x1)(x1-x3)(x3-x2) + x1(y3^2-y2^2) + x2(y1^2-y3^2) + x3(y2^2-y1^2)]/[x1(y3-y2) +x2(y1-y3) + x3(y2-y1)]


So, as we see, the formula for the centre of the circle coordinates exists, and the expression is symmetrical concerning the substitution of x to y coordinates of the points, and also symmetrical concerning the substitution of (x1,y1), (x2,y2) and (x3,y3) between each other.

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