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Question Number 213573 by mr W last updated on 09/Nov/24

Commented by mr W last updated on 09/Nov/24

$$findr=?$$

Commented by ajfour last updated on 09/Nov/24

Sir, this we hsd solved before.

Commented by mr W last updated on 09/Nov/24

$$i^{\prime }veforgottensir.$$

Answered by mr W last updated on 09/Nov/24

Commented by mr W last updated on 09/Nov/24

$$R=\frac{\sqrt{a^2+b^2-2ab\cos \alpha }}{2\sin \alpha }$$$$P(\frac{a}{2},\sqrt{R^2-\frac{a^2}{4}})$$$$Q(\frac{r}{\tan \frac{\alpha }{2}},r)$$$$PQ=R-r$$$$\sqrt{{(\frac{a}{2}-\frac{r}{\tan \frac{\alpha }{2}})}^2+{(\sqrt{R^2-\frac{a^2}{4}}-r)}^2}=R-r$$$$\frac{r}{{\tan }^2\frac{\alpha }{2}}=\frac{a}{\tan \frac{\alpha }{2}}+2(\sqrt{R^2-\frac{a^2}{4}}-R)$$$$\Rightarrow r=a\tan \frac{\alpha }{2}+2(\sqrt{R^2-\frac{a^2}{4}}-R){\tan }^2\frac{\alpha }{2}$$

Commented by ajfour last updated on 10/Nov/24

$$Ifsmallercircleissmall,then$$$${(R-r)}^2={(\frac{r}{\tan \frac{\alpha }{2}}-\frac{a}{2})}^2+{(r+\sqrt{R^2-\frac{a^2}{4}})}^2$$$$\Rightarrow r=\tan {}^2\left(\frac{\alpha }{2}\right)\{\frac{a}{\tan \frac{\alpha }{2}-2}-2R-2\sqrt{R^2-\frac{a^2}{4}}\}$$

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