সমকোণী ত্রিভুজের পরিকেন্দ্র

The circumcenter of a right triangle is the midpoint of the hypotenuse.

To prove this, consider the triangle $ABO$, where: $$O=(0,0)$$ $$A=(2a,0)$$ $$B=(0,2b)$$ $$M=(a,b)$$ Notice that $M$ is the midpoint of the hypotenuse $AB$.

Clearly, $MO=MA=MB=\sqrt{a^2+b^2}$. Thus, $M$ is equidistant from the vertices, so it is the circumcenter of $OAB$.

Alternatively, the perpendicular bisector of $OA$ is the line $x=a$, and the perpendicular bisector of $OB$ is the line $y=b$. These two intersect at $M$, which means that $M$ is the circumcenter of $OAB$.

Of course, there are also synthetic geometry proofs that employ roughly the same principle, but the coordinate method gives us very quick results in this case.