Серединні перпендикуляри перетинаються в центрі описаного кола

The circumcircle of a triangle is the unique circle going through the three vertices. The center of this circle is the circumcenter.

You may ask, where's the proof?

First, let's draw ABC.

I have shown the special case, of a $45-45-90$ triangle.

D,E,F are the midpoints of the sides, and the lines going through them are perpendicular to the sides.

Notice that any point P on the line going through F satisfies $PA=PB$.

Any point P on the line going through D satisfies $PB=PC$

Any point P on the line going through E satisfies $PA=PC$

Now we want to show that the perpendicular bisectors always intersect at one point.

Consider the medial triangle of ABC (which is DEF). It consists of drawing the triangle with the midpoints of ABC as vertices.

Notice that the perpendicular bisectors of $AB, BC, AC$ are the altitudes of triangle $DEF$.

Therefore, since the altitudes of a triangle are concurrent (we proved this in a different topic), the perpendicular bisectors of ABC are concurrent.

The circumradius is the radii length of the circumcircle. In any triange ABC, if $x$ is the area, then $x = \frac{abc}{4R}$ (R is the circumradius)

Therefore, if we know the area and side lengths, we can find R easily.

Another interesting thing: If R = circumradius and r = inradius, then $R \ge 2r$. I challenge you to prove this on your own.