In fact, triangles OBA and OB'A' are similar etc....

Consider now a triangle ABC, its circumcircle f and the inversion on a circle c' with arbitrary radius and center O on F. f transforms to a line e and points A, B, C respectively to points A', B', C' on that line. Start from the vector equation A'B'+B'C'+C'A' = 0. Divide with the distance d of O from e to obtain A'B'/d + B'C'/d + C'D'd = 0. But using the previous exercise |A'B'|/d=|AB|/z, |B'C'|/d = |BC|/x, |C'A'|/d = |CA|/y. Hence substituting in the previous equation we get:

a/x + b/y + c/z = 0.

Where a=|BC|, b=|CA|, c=|AB|. This is the equation of the circumcircle f of ABC in trilinear coordinates w.r. to that triangle. Since the barycentric coordinates (x',y',z') are equal to (x*a,y*b,z*c), we get the equation of the circumcircle in barycentric coordinates:

a

BarycentricsFormulas.html

BarycentricCoordinates.html

BarycentricCoordinates2.html

BarycentricCoordinates3.html

Isogonal_Conjugation.html

IsotomicConicOfLine.html

LineInTrilinears.html

Symmedian.html

TrilinearPolar.html

Trilinear_Polar.html

Tripole.html

Yiu, P.

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