3. The point on the circle $x^{2}+y^{2}=4$ that is closest to the line $4 x+3 y-12$ $=0$ has coordinates ( ).
(A) $\left(\frac{6}{5}, \frac{8}{5}\right)$
(B) $\left(\frac{8}{5}, \frac{6}{5}\right)$
(C) $\left(\frac{12}{5}, \frac{16}{5}\right)$
(D) $\left(\frac{16}{5}, \frac{12}{5}\right)$
3. B.

It is known that $A(3,0), B(0,4)$. As shown in Figure 1, the point with the minimum distance should lie on the tangent line parallel to the given line. Connecting the radius $O P$, we have $O P \perp A B$, thus
$\triangle P Q O \backsim \triangle A O B$.
Therefore, $\frac{x}{O P}=\frac{O B}{A B} \Rightarrow x=\frac{O P \cdot O B}{A B}=\frac{2 \times 4}{5}=\frac{8}{5}$.
$\frac{y}{O P}=\frac{O A}{A B} \Rightarrow y=\frac{O P \cdot O A}{A B}=\frac{2 \times 3}{5}=\frac{6}{5}$.