The equation to the sides $AB,BC,CA$ of a $\triangle$ are $x+y=1;4x-y+4=$ and $2x+3y=6$. Circle are drawn on $AB,BC,CA$ as diameter. The point of concurrence of the common chord is

The family of circles having the sides of a triangle as diameters has a specific radical axis property. According to the radical axis theorem for such a family, the pairwise common chords of these three circles are the altitudes of the triangle. Because the altitudes of any triangle intersect at a single point, the point of concurrence of these common chords is the orthocenter.

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