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Question Number 75684 by mr W last updated on 15/Dec/19

Commented by mr W last updated on 15/Dec/19

$$ifthesidelengthofsquareis1.$$$$findtheradiusofthesmallgreen$$$$circle.$$

Commented by vishalbhardwaj last updated on 15/Dec/19

$$\mathrm{sir}\mathrm{please}\mathrm{give}\mathrm{some}\mathrm{hint}\mathrm{to}\mathrm{solve}$$

Answered by MJS last updated on 15/Dec/19

$$\mathrm{the}\mathrm{idea}\mathrm{is},\mathrm{find}\mathrm{a}\mathrm{circle}\mathrm{which}\mathrm{touches}\mathrm{the}$$$$3\mathrm{given}\mathrm{circles}\mathrm{in}\mathrm{exactly}\mathrm{one}\mathrm{point}\mathrm{each}.$$$$\mathrm{I}\mathrm{get}$$$$r=\frac{4}{33}$$$$\mathrm{if}\mathrm{the}\mathrm{left}\mathrm{bottom}\mathrm{vertice}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{is}\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{green}\mathrm{circle}\mathrm{is}$$$${(x-\frac{20}{33})}^2+{(y-\frac{7}{11})}^2={\left(\frac{4}{33}\right)}^2$$$$\mathrm{or}$$$$y=\frac{7}{11}\pm \frac{\sqrt{-(11x-8)(33x-16)}}{11\sqrt{3}}$$

Commented by mr W last updated on 15/Dec/19

$$thankssir!i^{\prime }lltrytoseeificould$$$$getthesameresult.$$

Answered by mr W last updated on 15/Dec/19

Commented by mr W last updated on 15/Dec/19

$$A(0,0)$$$$C(x,y)$$$$x^2+y^2={(1-r)}^2...\left(i\right)$$$$x^2+{(y-\frac{1}{2})}^2={(\frac{1}{2}+r)}^2...\left(ii\right)$$$${(x-\frac{1}{2})}^2+{(1-y)}^2={(\frac{1}{2}-r)}^2...\left(iii\right)$$$$\left(i\right)-\left(ii\right):$$$$\frac{1}{2}(2y-\frac{1}{2})=\frac{3}{2}(\frac{1}{2}-2r)$$$$\Rightarrow y=1-3r$$$$\left(i\right)-\left(iii\right):$$$$\frac{1}{2}(2x-\frac{1}{2})+(2y-1)=\frac{1}{2}(\frac{3}{2}-2r)$$$$x+2y=2-r$$$$x+2(1-3r)=2-r$$$$\Rightarrow x=5r$$$${\left(5r\right)}^2+{(1-3r)}^2={(1-r)}^2$$$$33r^2=4r$$$$\Rightarrow r=\frac{4}{33}$$$$\Rightarrow y=1-\frac{3\times 4}{33}=\frac{7}{11}$$$$\Rightarrow x=5\times \frac{4}{33}=\frac{20}{33}$$$$\Rightarrow C(\frac{20}{33},\frac{7}{11})$$

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