6. Two vectors $\overrightarrow{O A}, \overrightarrow{O B}$ on a plane satisfy $|\overrightarrow{O A}|=a,|\overrightarrow{O B}|=b$, and $a^{2}+b^{2}=4, \overrightarrow{O A} \cdot \overrightarrow{O B}=0$. If the vector $\overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R})$, and $\left(\lambda-\frac{1}{2}\right)^{2} a^{2}+\left(\mu-\frac{1}{2}\right)^{2} b^{2}=1$, then the maximum value of $|\overrightarrow{O C}|$ is . $\qquad$
6.2 Analysis: Since $|\overrightarrow{O A}|=a,|\overrightarrow{O B}|=b$, and $a^{2}+b^{2}=4, \overrightarrow{O A} \perp \overrightarrow{O B}$, points $O, A, B$ lie on a circle with the midpoint $M$ of $A B$ as the center and 1 as the radius. Also, $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B}), \overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}$, so $\overrightarrow{M C}=\left(\lambda-\frac{1}{2}\right) \overrightarrow{O A}+$ $\left(\mu-\frac{1}{2}\right) \overrightarrow{O B}$, hence $|\overrightarrow{M C}|^{2}=\left(\lambda-\frac{1}{2}\right)^{2} a^{2}+\left(\mu-\frac{1}{2}\right)^{2} b^{2}=1$, thus, point $C$ also lies on the circle with $M$ as the center and 1 as the radius. Therefore, points $O, A, B, C$ are concyclic, with $M$ as the center. When $O, M, C$ are collinear, i.e., $O C$ is a diameter of $\odot M$, $|\overrightarrow{O C}|_{\max }=2$