10.014. A chord of a circle is equal to $10 \mathrm{~cm}$. Through one end of the chord, a tangent to the circle is drawn, and through the other end, a secant parallel to the tangent is drawn. Determine the radius of the circle if the inner segment of the secant is $12 \mathrm{~cm}$.
## Solution.

Draw the radius to point $A$; since $O A \perp A M$, then $O A \perp B C$ and $B D=6$ cm (Fig. 10.14). Let $O A=r, O D=x$. Then $A D=\sqrt{A B^{2}-B D^{2}}=8$, i.e., $r+x=8$ (1). But $O D^{2}=O B^{2}-B D^{2}$ or $r^{2}-x^{2}=36$; since

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-563.jpg?height=492&width=562&top_left_y=32&top_left_x=127)

Fig. 10.15

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-563.jpg?height=428&width=418&top_left_y=96&top_left_x=767)

Fig. 10.16

$r^{2}-x^{2}=(r+x)(r-x)$, then $r-x=36 / 8=9 / 2$

(2). Solving the system of equations (1) and (2), we find $r=6.25$ (cm).

Answer: $6.25 \text{ cm}$.