Let \(X\) be the point of contact of the smaller circle with the diameter, and let \(O_1\) be the centre of the smaller circle. Let the angles \(\angle O_1OC = \theta\) and \(\angle O_1AO = \alpha\).

The straight line \(OO_1\) meets the larger circle at its point of tangency with the smaller circle, and hence \(OO_1 = 1-r\). Thus \((1-r)\sin\theta = r\), and hence
\[ r \; = \; \frac{\sin\theta}{1 + \sin\theta} \]
Consequently
\[
AX \; = \; 1 + (1-r)\cos\theta \; = \; 1 + \frac{\cos\theta}{1+\sin\theta} \; = \; \frac{1+\sin\theta+\cos\theta}{1+\sin\theta} \; = \; \frac{2(1+\cos\theta)}{1+\sin\theta+\cos\theta} \]
the last identity being a trigonometric identity that is easy to prove. Hence
\[
AX \; =\; 2\left(1 - \frac{\sin\theta}{1+\sin\theta+\cos\theta}\right) \; = \; 2\left(1 - \frac{r}{AX}\right) \; = \; 2(1-\tan\alpha) \]
so that
\[ r \; = \; AX\tan\alpha \; = \; 2\big(\tan\alpha - \tan^2\alpha\big) \]
Since \(AB\) and \(AC\) are both tangents to the smaller circle we have \(\angle BAO_1 = \alpha\), and so it follows that \(\angle BAC = 2\alpha\), and hence \(a = 2\cos2\alpha\). Thus
\[ \frac{2-a}{2+a} = \frac{4\sin^2\alpha}{4\cos^2\alpha} \; = \; \tan^2\alpha \]
and hence we deduce that
\[
r \; = \; 2\left(\sqrt{\frac{2-a}{2+a}} - \frac{2-a}{2+a}\right)
\]
as required.
–
Mark Hennings
·
3 years, 10 months ago

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TopNewestLet \(X\) be the point of contact of the smaller circle with the diameter, and let \(O_1\) be the centre of the smaller circle. Let the angles \(\angle O_1OC = \theta\) and \(\angle O_1AO = \alpha\).

The straight line \(OO_1\) meets the larger circle at its point of tangency with the smaller circle, and hence \(OO_1 = 1-r\). Thus \((1-r)\sin\theta = r\), and hence \[ r \; = \; \frac{\sin\theta}{1 + \sin\theta} \] Consequently \[ AX \; = \; 1 + (1-r)\cos\theta \; = \; 1 + \frac{\cos\theta}{1+\sin\theta} \; = \; \frac{1+\sin\theta+\cos\theta}{1+\sin\theta} \; = \; \frac{2(1+\cos\theta)}{1+\sin\theta+\cos\theta} \] the last identity being a trigonometric identity that is easy to prove. Hence \[ AX \; =\; 2\left(1 - \frac{\sin\theta}{1+\sin\theta+\cos\theta}\right) \; = \; 2\left(1 - \frac{r}{AX}\right) \; = \; 2(1-\tan\alpha) \] so that \[ r \; = \; AX\tan\alpha \; = \; 2\big(\tan\alpha - \tan^2\alpha\big) \] Since \(AB\) and \(AC\) are both tangents to the smaller circle we have \(\angle BAO_1 = \alpha\), and so it follows that \(\angle BAC = 2\alpha\), and hence \(a = 2\cos2\alpha\). Thus \[ \frac{2-a}{2+a} = \frac{4\sin^2\alpha}{4\cos^2\alpha} \; = \; \tan^2\alpha \] and hence we deduce that \[ r \; = \; 2\left(\sqrt{\frac{2-a}{2+a}} - \frac{2-a}{2+a}\right) \] as required. – Mark Hennings · 3 years, 10 months ago

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