Area-preserving circle slice length

Question

I have a circle with a radius r and center O (1). It gets sliced. The length of the line from center O to the slice edge at 90° angle is know and is equal to r₂ (2). In this case I can easily figure out how to calculate the length l of the slice line. However, this won't work for me, because this is really a "slice" case and the circle doesn't preserve area. I, however, need to calculate the length of the line if it preserves the area. Now we get 2 lines(AB and CD) which both belog to corresponding tangents of circle(going through points A and C). AB and CD have identic length and they're placed as seen in the image (3). Now I need to calculate the l₂ created in this case. This is something that I, however, don't know how to calculate. What is an effective way to calculate this given arguments r and r₂(as far as I know, the rest can be calculated out of those 2 values)?

Answer 1

Referring to the figure shown below, by “to have the area preserved”, I guess you mean:-

[The cut-off segment QXR] is compensated by $2 \times$ [the brown shaded region].

The question is to find x, the distance of P from Q on the extension of MQ, such that [the brown shaded region] = [the light green shaded region].

Or equivalently, find x such that [quadrilateral ATPM] = [sector ATQX] …… (0)

In the following, all angles are measure in radians.

In $\triangle AMQ$, $s = \sqrt {r^2 – d^2}$ and $\alpha = \sin^{-1} (\dfrac {d}{r})$. Both of them are then known quantities.

In $\triangle AMP$, if we let $\angle APM = \beta$, then $2t = \dfrac {d}{\sin \beta}$ …... (1)

In $\triangle APT$, if we let $\angle APT = \phi$, then $\phi = \sin^{-1}(\dfrac {r}{2t})$ …... (2)

Also, $PT = u = \sqrt{(2t)^2 – r^2}$. After putting (1) in, we have $u = \sqrt{(\dfrac {d}{\sin \beta})^2 – r^2}$. …... (3)

Since $$[\text {quadrilateral ATPM}] = \dfrac {ru}{2} +\dfrac {d(x + s)}{2} = \dfrac {r\sqrt{(\dfrac {d}{\sin \beta})^2 – r^2}}{2} + \dfrac {d(x + s)}{2}$$

And since ATPM is cyclic, $$[\text {sector ATQX}] = \dfrac {r^2}{2} [\pi – ((\phi + \beta))]^c = \dfrac {r^2}{2} [\pi – ((\sin^{-1}(\dfrac {r}{2t}) + \beta))]^c = \dfrac {r^2}{2} [\pi – ((\sin^{-1}(\dfrac {r\sin\beta}{d}) + \beta))]^c $$

In short, from (0), we have $\dfrac {r\sqrt{(\dfrac {d}{\sin \beta})^2 – r^2}}{2} + \dfrac {d(x + s)}{2} = \dfrac {r^2}{2} [\pi – ((\sin^{-1}(\dfrac {r \sin \beta}{d}) + \beta))]$ …(4).

In addition, by surveyor’s formula, we have $\dfrac {x}{d} = \dfrac {1}{\tan \beta} - \dfrac {1}{\tan \alpha}$ …. (5)

Finally, the value of x (and also $\beta$) can be found theoretically but not readily from (4) and (5) because the two equations seem to be independent.