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Question Number 125782 by ajfour last updated on 13/Dec/20

Commented by ajfour last updated on 13/Dec/20

$$Findradiusofthesmallercircle.$$

Commented by MJS_new last updated on 14/Dec/20

$$\mathrm{sorry}\mathrm{I}\mathrm{have}\mathrm{no}\mathrm{time}...$$$$\mathrm{can}\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{all}\mathrm{circles}$$$${(x-m)}^2+{(y-n)}^2=n^2\mathrm{with}0⩽m⩽a$$$$\mathrm{touching}\mathrm{the}\mathrm{upper}\mathrm{half}\mathrm{ellipse}$$$$y=\frac{1}{a}\sqrt{a^2-x^2}\mathrm{with}1<a$$$$\mathrm{as}\mathrm{a}\mathrm{first}\mathrm{step}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{problem}?$$

Answered by mr W last updated on 14/Dec/20

$$let\mu =\frac{b}{a},\lambda =\frac{r}{a}$$$$sayP(a\cos \theta ,b\sin \theta )$$$$\tan \phi =-\frac{b\cos \theta }{-a\sin \theta }=\frac{\mu }{\tan \theta }$$$$a\cos \theta =\sqrt{{(\frac{b}{2}+r)}^2-{(\frac{b}{2}-r)}^2}+r\sin \phi$$$$\Rightarrow a\cos \theta =\sqrt{2br}+\frac{\mu r}{\sqrt{{\mu }^2+{\tan }^2\theta }}...\left(i\right)$$$$b\sin \theta =r+r\cos \phi$$$$\Rightarrow b\sin \theta =r[1+\frac{\tan \theta }{\sqrt{{\mu }^2+{\tan }^2\theta }}]...\left(ii\right)$$$$\left(ii\right)\mathbf{÷}\left(i\right):$$$$\mu \tan \theta =\frac{r[1+\frac{\tan \theta }{\sqrt{{\mu }^2+{\tan }^2\theta }}]}{\sqrt{2br}+\frac{r\mu }{\sqrt{{\mu }^2+{\tan }^2\theta }}}$$$$lett={\tan }^2\theta$$$$\mu t=\frac{\sqrt{\lambda }[1+\frac{\mathrm{t}}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]}{\sqrt{2\mu }+\frac{\sqrt{\lambda }\mu }{\sqrt{{\mu }^2+{\mathrm{t}}^2}}}$$$$\sqrt{\lambda }[1+\frac{\mathrm{t}}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]=\frac{\sqrt{\lambda }{\mu }^{2}t}{\sqrt{{\mu }^2+t^2}}+\mu t\sqrt{2\mu }$$$$\Rightarrow \lambda =\frac{2{\mu }^3{t}^2}{{[1+\frac{(1-{\mu }^2)t}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]}^2}...\left(I\right)$$$$from\left(ii\right):$$$$\frac{\mu t}{\sqrt{1+t^2}}=\lambda [1+\frac{\mathrm{t}}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]$$$$\Rightarrow \lambda =\frac{\mu t}{\sqrt{1+t^2}[1+\frac{\mathrm{t}}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]}...\left(II\right)$$$$\Rightarrow {[1+\frac{(1-{\mu }^2)t}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]}^2=2{\mu }^{2}t\sqrt{1+t^2}[1+\frac{\mathrm{t}}{\sqrt{{\mu }^2+{\mathrm{t}}^2}}]...\left(III\right)$$$$wecangettfrom\left(III\right)andthen\lambda$$$$from\left(I\right)or\left(II\right).$$

Commented by mr W last updated on 14/Dec/20

Commented by mr W last updated on 14/Dec/20

Commented by mr W last updated on 14/Dec/20

Commented by ajfour last updated on 14/Dec/20

$$Perfectlywellmanaged,Sir!$$$$Verypragmaticandwise{sol}^n.$$

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