4. The minimum value of the sum of the squares of the distances from a point in the rectangular coordinate plane to the three lines $x=0, y=0, 4x+3y=12$ is $\qquad$
(Tip: $x^{2}+y^{2}+\left(\frac{4 x+3 y-12}{\sqrt{3^{2}+4^{2}}}\right)^{2}=\frac{1}{2}\left[x^{2}+y^{2}\right.$
$$
\begin{array}{l}
\left.+\left(\frac{4 x+3 y-12}{5}\right)^{2}\right] \times\left[\left(-\frac{4}{5}\right)^{2}+\left(-\frac{3}{5}\right)^{2}+1\right] \geqslant \\
\left.\frac{1}{2}\left(-\frac{4}{5} x-\frac{3}{5} y+\frac{4}{5} x+\frac{3}{5} y-\frac{12}{5}\right)^{2}=\frac{72}{25} .\right)
\end{array}
$$