# Find the area of a circle that is inscribed in a circular sector with a radius $R$ and an angle $2x$.

The circle within the sector touches the radii R and the arc. So what is the area of the inscribed circle? The answer is actually $$S = \pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$ How can I derive this?

Call $\;O\;$ the center of the big circle from where the circular sector is taken, and let $\;M\;$ be the little circle's center with radius $\;r\;$, and $\;A,B\;$ the two tangent points between circle $\;M\;$ and the the two radii formining the sector.

If we put $\;t:=$the distance between $\;O\;$ and the inner circle, then taking the straight angle triangle $\;\Delta AMO\;$ we get:

$$\sin x=\frac r{r+t}\;\;\text{and also}\;\;r=R-(t+r)\implies t=R-2r\implies$$

$$r=(r+R-2r)\sin x\implies r=(R-r)\sin x\implies r=\frac{R\sin x}{1+\sin x}$$

so finally

$$S=\pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$

• Let's say that the tangent point between the inscribed circle and the arc of the big circle is T. I've noticed that if we prove that T, M and O are in a straight line, then everything's really easy. How can I do that? And I don't actually understand your way of solving this problem. $$\sin x = \frac {AM}{OM} = \frac {r}{t}$$ Why did you write $r+t$ and not $t$ instead of OM? I think you actually meant $t$ and not $r+t$ when you wrote that. Commented Nov 14, 2013 at 20:23
• If I'm correct that you wrote $r+t$ instead of $t$ by accident, then from what I see later on in the solution is that you implied that T, M and O are in the same straight line (where T is the tangent point between the little circle and the arc of the big circle). How do you know that and how can you prove it? Thanks. Commented Nov 14, 2013 at 20:23
• $\;OM=t+r=$the distance between $\;O\;$ and the inner, little circle + the little circle's radius. This is one of the main reasons why this kind of problems go way easier with adiagram with letters... Commented Nov 14, 2013 at 20:26
• Yes, there was a mistake in my answer. Again, it's hard not to get confused without a diagram. Commented Nov 14, 2013 at 20:27
• But do you know how to prove the later point that T, M and O are in the same straight line? Commented Nov 14, 2013 at 20:28

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$$\sin⁡x=\dfrac{r}{R-r}$$

$$⇒r=(R-r)sin⁡x$$

$$⇒r+r \sin⁡x= R \sin⁡x$$

$$⇒r(1+ \sin⁡x )=R \sin⁡x$$

$$⇒r^2 (1+ \sin⁡x )^2=R^2 \sin^2⁡x$$

$$⇒r^2=R^2 \dfrac{\sin^2⁡x}{(1+ \sin⁡x )^2}$$

$$⇒S=πr^2=πR^2 \dfrac{ \sin^2⁡x}{(1+ \sin⁡x )^2}$$

• Welcome to MSE. Try to use MathJax to format both your answers and questions :) Commented Sep 24, 2018 at 18:53
• thanks @mrtaurho Commented Sep 25, 2018 at 7:53