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, $2Rr_{A}=DI_{A}\cdot I_{A}A=OI_{A}^{2}-R^{2}$
Ortho means "straight, right". Orthocenter, because it is the intersection of the lines passing through the vertices and forming right-angles with the opposite sides.
There are many circumferences associated to the orthocenter. I don't think (or better said I haven't heard of) any of them called orthocircle.
Nevertheless, the orthocenter is the center of the circle that passes through the vertices of the anticomplementary triangle (it is the circumcenter of the anticomplementary triangle). The anticomplementary triangle has sides passing through the vertices of the triangle, parallel to the opposite sides.
Equivalently the original triangle has vertices at the midpoints of the sides of the anticomplementary triangle. Maybe this could be called the orthocircle.
Also, if you take a circle with its radius being half the circumradius of the triangle and with center halfway between the orthocentre and the circumcenter, we get the nine-point circle. This circle passes through the feet of the altitudes, the mid-points of the sides, and the mid-points between the orthocenter and the vertices.
In the picture above, $H$ is the orthocenter, $O$ is the circumcenter, and $G$ is the center of the nine-point circle, which is halfway between $H$ and $O$.
The nine-point circle is $1/4$ times contraction of the circumcenter of the anticomplementary triangle, again a contraction with center at the orthocenter. Equivalently, the circumcenter of the anticomplementary triangle is the $2$ times expansion of the circumcenter of the original triangle, by a contraction again with center at the orthocenter.
Best Answer
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?