# circle

please......can anyone solve it for me......i m in a need of help..............

Note by Ronak Pawar
4 years, 10 months ago

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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.

- 4 years, 9 months ago

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