5. In triangle $\mathrm{ABC}$, the sides $A B=4, B C=6$. Point $M$ lies on the perpendicular bisector of segment $A B$, and lines $A M$ and $A C$ are perpendicular. Find $M A$, if the radius of the circumscribed circle around triangle $A B C$ is 9.
Solution. Introduce a coordinate system with the origin at point A such that point C lies on the x-axis. From the problem statement, point M lies on the y-axis. Let's introduce notations for the unknown coordinates: $\mathrm{A}(0,0), \mathrm{B}\left(\mathrm{x}_{\mathrm{B}}, \mathrm{y}_{\mathrm{B}}\right), \mathrm{C}\left(\mathrm{x}_{\mathrm{C}}, 0\right), \mathrm{M}\left(0, \mathrm{y}_{\mathrm{M}}\right)$. Let $\mathrm{N}$ be the midpoint of $\mathrm{AB}$ and 0 be the center of the circumscribed circle, then $N\left(\frac{x_{B}}{2}, \frac{y_{B}}{2}\right)$ and $O\left(\frac{x_{C}}{2}, y_{0}\right)$. From the perpendicularity of vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{MN}}$, it follows that $\mathrm{x}_{\mathrm{B}} \frac{\mathrm{x}_{\mathrm{B}}}{2}+\mathrm{y}_{\mathrm{B}}\left(\mathrm{y}_{\mathrm{M}}-\frac{\mathrm{y}_{\mathrm{B}}}{2}\right)=0$. From this, considering $\mathrm{AB}^{2}=\mathrm{x}_{\mathrm{B}}^{2}+\mathrm{y}_{\mathrm{B}}^{2}=16$, we get $\mathrm{y}_{\mathrm{M}}=\frac{8}{\mathrm{y}_{\mathrm{b}}}$. From the perpendicularity of vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{MO}}$, it follows that $\mathrm{x}_{\mathrm{B}} \frac{\mathrm{x}_{C}}{2}+\mathrm{y}_{\mathrm{B}}\left(\mathrm{y}_{\mathrm{M}}-\mathrm{y}_{0}\right)=0$. Additionally, $\mathrm{BC}^{2}=\left(\mathrm{x}_{\mathrm{B}}-\mathrm{x}_{C}\right)^{2}+\mathrm{y}_{\mathrm{B}}^{2}=36$ and $\mathrm{AO}^{2}=\left(\frac{\mathrm{x}_{\mathrm{C}}}{2}\right)^{2}+\mathrm{y}_{0}^{2}=81$. These equations are sufficient to obtain $\mathrm{MA}=\left|\mathrm{y}_{\mathrm{M}}\right|=6$.

Answer: 6.