3A. On the line $x=3$ find a point $T$ such that the trapezoid formed by the two tangents drawn from point $T$ to the circle $x^{2}+y^{2}=9$ and the coordinate axes has the smallest area.
Solution. One of the tangents drawn from $T$ to the given circle has the equation $x=3$. The point $T$ has coordinates $\left(3, y_{0}\right)$. Let the equation of the second tangent be $y=k x+l$. Then we have

$$
9\left(1+k^{2}\right)=l^{2}
$$

Since the point $T\left(3, y_{0}\right)$ lies on $y=k x+l$, we get

$$
y_{0}=3 k+l
$$

From (1) and (2) it follows that

$$
k=\frac{y_{0}^{2}-9}{6 y_{0}}, \quad l=\frac{y_{0}^{2}+9}{2 y_{0}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_bd3f37f65d5d6e754f63g-12.jpg?height=342&width=450&top_left_y=1050&top_left_x=1015)

Now for the area of the trapezoid, we get

$$
P=\frac{3\left(y_{0}+l\right)}{2}=\frac{9}{4}\left(y_{0}+\frac{3}{y_{0}}\right) \geq \frac{9}{2} \sqrt{y_{0} \cdot \frac{3}{y_{0}}}=\frac{9 \sqrt{3}}{2}
$$

where the equality holds if and only if $y_{0}=\frac{3}{y_{0}}$, i.e., if and only if $y_{0}= \pm \sqrt{3}$.

Finally, the desired points are $T(3, \pm \sqrt{3})$, and the maximum area is $\frac{9}{2} \sqrt{3}$ square units.