$AOB$ 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 $AB$.
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Anything between $0$ and $5$. – Feb 19 '16 at 08:16
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I need a formula, not a forcast – Nguyen Viet Anh Feb 19 '16 at 08:17
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1$$0<r<5$$ if you prefer. – Feb 19 '16 at 08:17
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@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 08:18
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Sorry, I forgot to mention that we have length of OA, angle at O – Nguyen Viet Anh Feb 19 '16 at 08:21
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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 $$ \sin\left(\frac\theta2\right)=\frac{r}{R-r} $$
robjohn
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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.
Swapnil Rustagi
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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 08:28
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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 08:54
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centre of a circle, inscribed in an angle, is on the bisector of this angle, but how we prove that touchpoint of this circle with an arc lies on this bisector. – math boy Mar 26 '22 at 21:51