4. Through the moving point $M$, draw the tangent line $M N$ to the circle $C:(x-2)^{2}+(y-2)^{2}=1$, where $N$ is the point of tangency. If $|M N|=|M O|(O$ is the origin), then the minimum value of $|M N|$ is $\qquad$ .
Answer $\frac{7 \sqrt{2}}{8}$.
Analysis Let $M(x, y)$, from $|M N|=|M O|$ we know: $|M O|^{2}+|N C|^{2}=|M C|^{2}$, that is
$$
x^{2}+y^{2}+1=(x-2)^{2}+(y-2)^{2}
$$

Simplifying to: $4 x+4 y-7=0$, so the locus of point $M$ is a line, so the minimum value of $|M N|$ equals the minimum value of $|M O|$, which is the distance from point $O$ to the line $4 x+4 y-7=0$, $\frac{7 \sqrt{2}}{8}$.