Is there a formula describing the length of line subtended from a chord in a circle? (Picture Included)

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Is there a formula that describes the length of line x in a circle, assuming you know the chord length (c), perpendicular distance of the chord from the circumference (d), circle radius (r), and the angle that line x subtends with the chord (a).

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  1. Observe that $MO = r -d$ and since the triangle $AMO$ is right angled, $$AM =\frac{c}{2}= \sqrt{OA^2 - OM^2} = \sqrt{r^2 - (r-d)^2}$$

  2. Triangle $MNO$ is right angled and $\angle \, OMN = a$ so $$MN = MO \,\cos(a) = (r-d) \,\cos(a), \,\,\,\, ON = MO \,\sin(a) = (r-d)\, \sin(a)$$

  3. Triangle $OND$ is right angled so by Pythagoras' theorem $$ND = \sqrt{OD^2 - ON^2} = \sqrt{r^2 -(r-d)^2 \sin^2(a)}$$

  4. By the power of point relation \begin{align} \frac{c^2}{4} &= AM^2 = AM \cdot MB = CM \cdot MD = CM \cdot \Big(MN + ND\Big)\\ &= x \, \Big( (r-d) \,\cos(a) + \sqrt{r^2 -(r-d)^2 \sin^2(a)}\Big) \end{align} thus

$$x = \frac{c^2}{4 \, \Big( (r-d) \,\cos(a) + \sqrt{r^2 -(r-d)^2 \sin^2(a)}\Big)}$$ or equivalently, if you prefer $$x = \frac{r^2 - (r-d)^2}{(r-d) \,\cos(a) + \sqrt{r^2 -(r-d)^2 \sin^2(a)}}$$

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Triangle

We do not need $c$ to calculate $x$, since $r$ and $d$ determine $c$. Draw a radius to where the line of length $x$ meets the circle. Using the law of cosines we may calculate $x$ as the root of a quadratic equation: $$r^2=x^2+(r-d)^2-2x(r-d)\cos(\pi-a)$$ $$x^2+(2(r-d)\cos a)x+d^2-2rd=0$$ One root of this equation will always be negative and can be ignored.