429. From a point outside a given circle, draw a secant such that its inner part has a given length.

## Rayno's Problem.
429. In this circle (Fig. 94), draw a chord $M N$ of a given length. Describe a circle from the center of the given circle, tangent to $M N$, and from point $A$, draw a tangent $A C$ to this circle. This will be the required secant.

This method of solution is constructive.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-186.jpg?height=457&width=582&top_left_y=1782&top_left_x=1114)

Using algebra in geometry, the problem is solved as follows: let the radius of the circle be $r$, $A K = a$, and $C B = d$ (Fig. 95). Then, denoting the secant $A C$ by $x$, we have:

from which

$$
\begin{gathered}
x(x-d)=a(a+2 r) \\
x^{2}-d x-a(a+2 r)=0
\end{gathered}
$$

$$
x=\frac{d}{2} \pm \sqrt{\left(\frac{d}{2}\right)^{2}+a(a+2 r)}
$$

The construction is clear.

Reynaud, A., a French mathematician-pedagogue, author of several textbooks, who participated in the later processing of Bezout's courses.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-187.jpg?height=385&width=643&top_left_y=687&top_left_x=358)

Fig. 95.