The radical center of the three circles descr | vedclass

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(C) Let the vertices of the triangle be $A, B,$ and $C$. The equations of the circles with sides $BC, CA,$ and $AB$ as diameters are given by:$S_1: (x

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Orthocenter of the triangle

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(C) Let the vertices of the triangle be $A, B,$ and $C$. The equations of the circles with sides $BC, CA,$ and $AB$ as diameters are given by:$S_1: (x-x_B)(x-x_C) + (y-y_B)(y-y_C) = 0$$S_2: (x-x_C)(x-x_A) + (y-y_C)(y-y_A) = 0$$S_3: (x-x_A)(x-x_B) + (y-y_A)(y-y_B) = 0$The radical axis of $S_1$ and $S_2$ is $S_1 - S_2 = 0$,which simplifies to the equation of the altitude from $C$ to $AB$.Similarly,the radical axis of $S_2$ and $S_3$ is the altitude from $A$ to $BC$.The intersection of these radical axes is the point where all three altitudes meet,which is the orthocenter of the triangle.

The radical center of the three circles described with the three sides of a triangle as diameters is:

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