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Question Number 166208 by ajfour last updated on 15/Feb/22

Commented by ajfour last updated on 15/Feb/22

$$Ifbothcircleshaveequalradii,$$$$findit.$$

Answered by Eulerian last updated on 15/Feb/22

$$\overline{\mathrm{AB}}=\sqrt{{\mathrm{r}}^2+{4\mathrm{r}}^2}=\sqrt{5}\mathrm{r}$$$$\mathrm{\angle }\mathrm{BAC}={\tan }^{-1}\left(\frac{\mathrm{r}}{2\mathrm{r}}\right)+{\cos }^{-1}\left(\frac{\mathrm{r}}{2\mathrm{r}}\right)={\tan }^{-1}\left(\frac{1}{2}\right)+{\cos }^{-1}\left(\frac{1}{2}\right)$$$$\therefore \mathrm{By}\mathrm{pythagorean}\mathrm{identity},\mathrm{we}\mathrm{have}$$$$\frac{\overline{\mathrm{BC}}}{\sqrt{5}\mathrm{r}}=\sin ({\tan }^{-1}\left(\frac{1}{2}\right)+{\cos }^{-1}\left(\frac{1}{2}\right))$$$$=\sin \left({\tan }^{-1}\left(\frac{1}{2}\right)\right)\cdot \cos \left({\cos }^{-1}\left(\frac{1}{2}\right)\right)+\cos \left({\tan }^{-1}\left(\frac{1}{2}\right)\right)\cdot \sin \left({\cos }^{-1}\left(\frac{1}{2}\right)\right)$$$$=\left(\frac{\sqrt{5}}{5}\right)\left(\frac{1}{2}\right)+\left(\frac{2\sqrt{5}}{5}\right)\left(\frac{\sqrt{3}}{2}\right)$$$$=\frac{\sqrt{5}+2\sqrt{15}}{10}$$$$\therefore$$$$\overline{\mathrm{BC}}=\left(\frac{1+2\sqrt{3}}{2}\right)\cdot \mathrm{r}$$$$$$$$\mathrm{Thus},\mathrm{we}\mathrm{now}\mathrm{have}:$$$$\left(\frac{1+2\sqrt{3}}{2}\right)\cdot \mathrm{r}+\mathrm{r}={\mathrm{r}}^2$$$$\mathrm{r}=\frac{1+2\sqrt{3}}{2}+1=\frac{3+2\sqrt{3}}{2}\mathrm{units}$$

Commented by ajfour last updated on 15/Feb/22

$$BC\ne \left(\sqrt{5}r\right)\sin \left(\mathrm{\angle }BAC\right)$$$$pleasecheck!sir.$$

Answered by mr W last updated on 15/Feb/22

Commented by Tawa11 last updated on 16/Feb/22

$$\mathrm{Great}\mathrm{sir}.$$

Commented by mr W last updated on 16/Feb/22

$$P(p,p^2)$$$$\tan \theta =2p$$$$x_A=p+r\sin \theta$$$$y_A=p^2-r\cos \theta$$$$Q(q,q^2)$$$$\tan \phi =2q$$$$x_B=q+r\sin \phi$$$$y_B=q^2-r\cos \phi$$$$$$$$x_B=x_A+r$$$$q+\frac{2qr}{\sqrt{1+4q^2}}=p+(1+\frac{2p}{\sqrt{1+4p^2}})r$$$$q-p=(1+\frac{2p}{\sqrt{1+4p^2}}-\frac{2q}{\sqrt{1+4q^2}})r...\left(i\right)$$$$$$$${(y_B-y_A)}^2={\left(2r\right)}^2-r^2$$$$y_B-y_A=\sqrt{3}r$$$$q^2-\frac{r}{\sqrt{1+4q^2}}-p^2+\frac{r}{\sqrt{1+4p^2}}=\sqrt{3}r$$$$q^2-p^2=(\sqrt{3}-\frac{1}{\sqrt{1+4p^2}}+\frac{1}{\sqrt{1+4q^2}})r...\left(ii\right)$$$$$$$$y_A=r$$$$p^2-r\cos \theta =r$$$$r=\frac{p^2}{1+\frac{1}{\sqrt{1+4p^2}}}...\left(iii\right)$$$$$$$$thereisonlyonesolution:$$$$p\approx 0.71681$$$$q\approx 1.01832$$$$r\approx 0.3268$$

Commented by mr W last updated on 16/Feb/22

Commented by mr W last updated on 16/Feb/22

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