4. Given a circle centered at the origin with radius $R(R>0)$ intersects the line segment connecting $M(1,1)$ and $N\left(\frac{7}{4}, 0\right)$. Then the range of values for $R$ is $\qquad$.
4. $\left[\frac{7}{5}, \frac{7}{4}\right]$.

Let the equation of $M N$ be $4 x+3 y-7=0$. Then the line passing through the origin and perpendicular to $M N$ is $3 x-4 y=0$. We can find the foot of the perpendicular from the origin to $M N$ to be $H\left(\frac{28}{25}, \frac{21}{25}\right)$, and
$$
O H=\frac{|4 \times 0+3 \times 0-7|}{\sqrt{4^{2}+3^{2}}}=\frac{7}{5} .
$$

Also, $O M=\sqrt{2}, O N=\frac{7}{4}$. Since $\frac{7}{5}<\sqrt{2}<\frac{7}{4}$, we have $O H<O M<O N$.
When and only when $O H \leqslant R \leqslant O N$, the circle intersects the line segment $M N$, giving the range of $R$ as $\left[\frac{7}{5}, \frac{7}{4}\right]$.