# Calculate the diameter of an inscribed circle inside a sector of circle [closed]

$AOB$ is a sector of a circle with center $O$, angle = 45° and radius $OA=10$. Find the radius of the chord inscribed circle in this sector such that it touches radius $OA$, radius $OB$ and arc $AB$. ## closed as off-topic by colormegone, Claude Leibovici, Em., Harish Chandra Rajpoot, user228113 Feb 19 '16 at 10:37

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• Anything between $0$ and $5$. – Yves Daoust Feb 19 '16 at 8:16
• I need a formula, not a forcast – Nguyen Viet Anh Feb 19 '16 at 8:17
• $$0<r<5$$ if you prefer. – Yves Daoust Feb 19 '16 at 8:17
• @NguyenVietAnh With the info given, "Between $0$ and $5$" is all we can say. We need something else, like the length of either the line segment $AB$ or the circle arc $AB$, or the angle at $O$. In other words, we need to know how big the sector is. – Arthur Feb 19 '16 at 8:18
• Sorry, I forgot to mention that we have length of OA, angle at O – Nguyen Viet Anh Feb 19 '16 at 8:21

## 2 Answers

The geometry in says $$r=(R-r)\sin\left(\frac\theta2\right)$$ Therefore, we can solve for $r$: $$\bbox[5px,border:2px solid #C0A000]{r=R\,\frac{\sin\left(\frac\theta2\right)}{1+\sin\left(\frac\theta2\right)}}$$ To find $\theta$ from $R$ and $r$, we can use $$\sin\left(\frac\theta2\right)=\frac{r}{R-r}$$

If $\theta$ is the sector angle, a coordinate geometry solution would be to consider a circle $x^2 + y^2 + 2gx + 2fy + c =0$ inside the circle $x^2 + y^2 = 100$. Now our required circle touches lines $y=0$, $y=\tan \theta x$ and the circle. I suppose you can do it now. For tangency to lines, distance of centre from line is equal to radius, add for touching circle, distance between centres = difference of radii.

• Poorly, I could not fully understand what you meant. If you do not mind, please write down a complete formula base on 2 input, length of OA and sector angle (for e.g. 10 and 45°) – Nguyen Viet Anh Feb 19 '16 at 8:28
• As I do not have any data, I am not sure, but try this: Radius of inscribed circle = $$\frac{R}{1+ \sqrt{(1+ \frac {\sec^2 \theta + 2 \sec \theta + 1}{\tan^2 \theta})}}$$ – Swapnil Rustagi Feb 19 '16 at 8:54