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?
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2 Answers
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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
$$S=\pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$
answered Nov 14, 2013 at 19:07
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$\begingroup$ 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. $\endgroup$ Commented Nov 14, 2013 at 20:23
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$\begingroup$ 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. $\endgroup$ Commented Nov 14, 2013 at 20:23
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$\begingroup$ $\;OM=t+r=$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... $\endgroup$ Commented Nov 14, 2013 at 20:26
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$\begingroup$ Yes, there was a mistake in my answer. Again, it's hard not to get confused without a diagram. $\endgroup$ Commented Nov 14, 2013 at 20:27
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$\begingroup$ But do you know how to prove the later point that T, M and O are in the same straight line? $\endgroup$ Commented Nov 14, 2013 at 20:28
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$\sinx=\dfrac{r}{R-r}$
$⇒r=(R-r)sinx$
$⇒r+r \sinx= R \sinx$
$⇒r(1+ \sinx )=R \sinx$
$⇒r^2 (1+ \sinx )^2=R^2 \sin^2x$
$⇒r^2=R^2 \dfrac{\sin^2x}{(1+ \sinx )^2} $
$⇒S=πr^2=πR^2 \dfrac{ \sin^2x}{(1+ \sinx )^2}$
