There is a circle $k$ and a point $P$ outside it. A secant $s$ passing through $P$ intersects $k$ at points $B$ and $C$, with $B$ on the segment $PC$. Construct the position of $s$ such that $BC \cdot PC = d^2$, where $d$ is a given segment.
According to a known theorem,

$$
P B \cdot P C = P E^{2}
$$

where $E$ is the point of tangency of the tangent drawn from $P$ to $k$. On the other hand, the requirement is

$$
B C \cdot P C = d^{2}
$$

The sum of (1) and (2) is

$$
(P B + B C) P C = P C^{2} = P E^{2} + d^{2}
$$

According to this, the distance of the sought position $s$ from $P$, which is the farther intersection point of the circle with $P$, can be constructed based on the Pythagorean theorem, see the triangle $P E D$ in the figure. Point $C$ is determined by the auxiliary circle drawn around $P$ with radius $P D$ intersecting $k$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b790bc5f258b507aa09ag-1.jpg?height=583&width=672&top_left_y=690&top_left_x=716)

The number of solutions depends on the size relationship between $P D$ and $P F$, where $F$ is the farthest point of $k$ from $P$, specifically 2, 1, or 0, depending on whether $P D \lesseqgtr P F$. (Note: $P D > P E > P G$.)

It is easy to show that the secants $s_{1}, s_{2}$ formed with the points $C_{1}, C_{2}$ satisfy the requirement.

The solvability and the number of solutions can also be formulated in terms of the original data: the radius $r$ of $k$, the distance $t$ from $P$ to the center $O$ of the circle $k$ ($t > r$), and $d$. Indeed, $P E^{2} = t^{2} - r^{2}$ and $P F = t + r$, and thus there are 2, 1, or 0 solutions, depending on whether $d^{2} \lesseqgtr 2 r(t + r)$.

András Benczur (Budapest, Fazekas M. Gimn. II. o. t.)