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On a circle with known radius ($x$ units). Along the diameter of the circle, there is a point with known distance ($y$ units) from the circumference (along the diameter). If this point subtends an angle $\theta$ on the circumference, how can I find the area of the resultant smaller sector?

Thanks

enter image description here

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  • $\begingroup$ could you post a picture? $\endgroup$ – Arnaldo Jan 26 '17 at 15:40
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    $\begingroup$ Hint: that's a circular sector minus two triangles. $\endgroup$ – dxiv Jan 26 '17 at 17:23
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Since the area you want to calculate depends on the radius and $y$ can be chosen independently of the radius, you also need to know the radius.


I created a slightly modified version of your picture, so I am sure about the notation. If I get you right, then $y=JI$, $\theta=\alpha$. Further I will call the radius $r$ and $x=AJ$.

enter image description here

The area $A$ you are interested in is composed of two parts whose areas are easy to calculate. The triangle $\Delta FIE$ and the rest of the area called $X$.

The triangle's area is two times the area of the right triangle $\Delta JIE$, so $A(\Delta FIE)=2 A(\Delta JIE)=2 y^2 \cot(\alpha)$

The other area is more complicated. The idea is to regard it as the circular sector spanned by the radii $AF$ and $AI$, minus the triangle $\Delta FIA$ spanned by these segments. Then $A(X)=\pi r^2 \frac {\beta} \pi - x y$.

We still need to express $x$ and $\beta$ in dependence of $r,y,\alpha$.

Therefore we look at the right triangle $\Delta JIA$: First, $\sin(\beta)=\frac y r$. Secondly, $x=\sqrt{r^2-y^2}$

Hence in conclusion $$A=2 y^2 \cot(\alpha) +r^2 \arcsin \left(\frac y r\right)-y \sqrt{r^2-y^2}.$$

In particular, if $y=k r$, then $$A=r^2 (2 k^2 \cot(\alpha) + \arcsin(k)-k \sqrt{1-k^2}).$$

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