Let $\Delta ABC$ be our triangle, $\Phi$ be another circle and let $O$ be a center of the circle.
Thus, $OA=OB$, which says that $O$ is placed on the perpendicular bisector to $AB$.
By the same way we obtain that $O$ is placed on the perpendicular bisector to $AC$.
But, you said that you know that these perpendicular bisectors intersects in your center of your circle, say $T$.
But since two different lines have an unique common point, we obtain $O\equiv T$.
Can you end it now?
First, even though this isn't what you asked, I wanted to explain why the 3 perpendicular bisectors are concurrent in the first place, and why they intersect at the circumcenter.
Basically, in order to circumscribe a triangle using a circle, you need to find the point that is equidistant from each of the vertices in the triangle. A perpendicular bisector of a segment AB is actually the set of all points that are equidistant from A and B. You can see this by picking an arbitrary point C on the line, and the intersection of the perpendicular bisector of AB with AB D. It is easy to see that AD=DB, CD=CD, and m<ADC=<BDC=90ยบ. Therefore, triangles ADC and BDC are congruent by SAS, so AC=BC.
Now let's focus on the triangle in question. Call it XYZ. Call line z the perpendicular bisector of XY, y the bisector of XZ, and x the bisector of YZ. The intersection of x and z is the point equidistant from X, Y, and Z, and since two lines can only intersect at one point, there is only one point with this property, This means that, since x and y also intersect at a point with this property, and the only point with that property is the intersection of x and z, we know that x, y, and z must be concurrent at the point equidistant from X, Y, and Z, called the circumcenter of triangle XYZ.
So, yes, you can find the circumcenter using only two perpendicular bisectors, but using 3 is usually a good thing to do since if they don't concur at the same point you know you made a mistake :)
Best Answer
angle bisector $AI$ cut the circumcircle of $\triangle ABC$ at $D$
$\angle DBI=\angle DBC+\angle IBC=\angle DAB+\angle ABI=\angle BID$ and then $DB=DI$
Likewise, $DC=DI$ and then $DB=BI=DC$
$I_{A}C$ bisect $\angle BCT$ $\Longrightarrow$ $\angle ICI_{A}=90^{\circ}$
thus, $DI=DC=DI_{A}$
the perpendicular bisector of $BC$ cut the circumcircle of $\triangle ABC$ at $M$
$\triangle SAI_{A}$ is similar to $\triangle BMD$
power of $I_{A}$ with respect to the circumcircle of $\triangle ABC$ is $OI_{A}^{2}-R^{2}$
Also, it is $I_{A}D\cdot I_{A}A$
$\triangle SAI_{A}\sim\triangle BMD$ $\Longrightarrow$ $\dfrac{MD}{BD}=\dfrac{I_{A}A}{SI_{A}}$
$\Longrightarrow$ $2Rr_{A}=BD\cdot I_{A}A=DI_{A}\cdot I_{A}A$
hence, $2Rr_{A}=DI_{A}\cdot I_{A}A=OI_{A}^{2}-R^{2}$