4B. At the point $A(4,0)$, a tangent to the circle $x^{2}+y^{2}=16$ is drawn. Determine the point $B$ on this tangent such that the area of the trapezoid formed by the coordinate axes and the tangents to the circle drawn from point B has an area of 26.
Solution. Let $t$ be the tangent line to the circle drawn at point $A$ and $B(4, b) \in t$. Let $C(0, c)$ denote the intersection point of the line $B C$ and the $y$-axis. Then for the area of the trapezoid $O A B C$ we get

$$
\frac{b+c}{2} \cdot 4=26 \text {, i.e. } b+c=13
$$

Furthermore, the equation of the line $B C$ is $y=\frac{b-c}{4} x+c$. The condition for the line

![](https://cdn.mathpix.com/cropped/2024_06_05_46a9829282eea0e99a44g-11.jpg?height=413&width=617&top_left_y=995&top_left_x=845)
$y=k x+n$ to be a tangent to the circle $x^{2}+y^{2}=r^{2}$ is $r^{2}\left(1+k^{2}\right)=n^{2}$, so we get $16\left(1+\left(\frac{b-c}{4}\right)^{2}\right)=c^{2}$ and from this $c=\frac{16+b^{2}}{2 b}$. Substituting $b+c=13$ into the last equation, we get $3 b^{2}-26 b+16=0$. The solutions to this equation are $b_{1}=\frac{2}{3}$ and $b_{2}=8$. The point $B$ can be on either side of the $x$-axis (the circle is symmetric with respect to the $x$-axis), so we get 4 points that satisfy the condition of the problem. They are $B_{1}(4,8), B_{2}(4,-8), B_{3}\left(4, \frac{2}{3}\right)$ and $B_{4}\left(4,-\frac{2}{3}\right)$.