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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply