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A circle has been built on each side of a 11-angled polygon(it may be not convex), where the side is the diameter of the circle. How can I prove that if all of these circles have an intersection point, the intersection point will coincide with one of the vertexes of the given polygon?

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    The circle with diameter $AB$ is the locus of points $P$ such that $\widehat{APB}$ is a right angle. If all those circle have a common point, such a point "sees" every side of the given 11-agon under a right angle. – 2017-02-26
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    And what does this fact about seeing the side under a right angle give us? – 2017-02-26
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    Think about it. Is there such point for a regular $11$-agon? – 2017-02-26
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    It does not, of course – 2017-02-26
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    Then consider my extended hint below: $4$ consecutive vertices distinct from $P$ make a whole "turn" around $P$, but there are $11$ vertices, and $11$ is not a multiple of $4$, so $P$ has to be one of the $11$ vertices. – 2017-02-26

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The circle with diameter $AB$ is the locus of points P such that $\widehat{APB}$ is a right angle. If all those circle have a common point, such a point "sees" every side of the given 11-agon under a right angle, or is an endpoint of such a side.

Extended hint: enter image description here

Is it possible that such a path closes in $11$ steps withouth going through $P$, since $4\nmid 11$?