3. In the Cartesian coordinate system $x O y$, two circles both pass through the point $(1,1)$, and are tangent to the line $y=\frac{4}{3} x$ and the $x$-axis. Then the sum of the radii of the two circles is . $\qquad$
3. $\frac{3}{2}$.

Let the equations of the two circles be
$$
\left(x-a_{i}\right)^{2}+\left(y-r_{i}\right)^{2}=r_{i}^{2}(i=1,2) \text {. }
$$

Since both circles pass through the point $P(1,1)$, we have
$$
\begin{array}{l}
\left(1-a_{i}\right)^{2}+\left(1-r_{i}\right)^{2}=r_{i}^{2} \\
\Rightarrow a_{i}^{2}-2 a_{i}-2 r_{i}+2=0 .
\end{array}
$$

Let the inclination angle of the line $y=\frac{4}{3} x$ be $\alpha$. Then
$$
\begin{array}{l}
\alpha \in(0, \pi), \tan \alpha=\frac{4}{3} \\
\Rightarrow \tan \frac{\alpha}{2}=\frac{1}{2} \Rightarrow \frac{r_{i}}{a_{i}}=\frac{1}{2} .
\end{array}
$$

Substituting into equation (1) and rearranging, we get $2 r_{i}^{2}-3 r_{i}+1=0$.
By Vieta's formulas, we know $r_{1}+r_{2}=\frac{3}{2}$.