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Question Number 43360 by ajfour last updated on 10/Sep/18

Commented by ajfour last updated on 10/Sep/18

$$Find\mathit{R}intermsof\mathit{a},\mathit{b},\mathit{c}.$$

Commented by ajfour last updated on 10/Sep/18

Commented by ajfour last updated on 10/Sep/18

$$pleasehelp..$$

Commented by MJS last updated on 10/Sep/18

$$\mathrm{same}\mathrm{as}\mathrm{question}37209$$

Commented by ajfour last updated on 10/Sep/18

$$ThanksagainSir,thesolution$$$$thereisnotlegible,pluspretty$$$$lengthyitis.Ihavenotunderstood$$$$solvingityet.$$

Commented by MJS last updated on 10/Sep/18

$$\mathrm{start}\mathrm{with}A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}0\\ a+b\end{array}\right)$$$$\mathrm{then}\mathrm{calculate}C=\left(\begin{array}{c}x\\ y\end{array}\right)$$$$\mid AC{\mid }^2={(a+c)}^2$$$$\mid BC{\mid }^2={(b+c)}^2$$$$\mathrm{this}\mathrm{can}\mathrm{be}\mathrm{solved}\mathrm{for}x\mathrm{and}y$$$$\mathrm{we}\mathrm{get}2\mathrm{solutions},\mathrm{one}\mathrm{with}x<0,\mathrm{the}\mathrm{other}$$$$\mathrm{with}x>0$$$$$$$$\mathrm{now}\mathrm{it}\mathrm{depends}\mathrm{on}c$$$$\left(1\right)\mathrm{if}c\mathrm{is}\mathrm{very}\mathrm{small},\mathrm{the}‘‘{\mathrm{circumcircle}}^{″}\mathrm{touches}$$$$\mathrm{the}\mathrm{three}\mathrm{circles}\mathrm{from}\mathrm{outside}$$$$\left(2\right){\mathrm{there}}^{\prime }\mathrm{s}\mathrm{exactly}\mathrm{one}\mathrm{value}\mathrm{of}c\mathrm{so}\mathrm{that}\mathrm{the}$$$$‘‘{\mathrm{circumcircle}}^{″}\mathrm{is}\mathrm{a}\mathrm{straight}\mathrm{line}=\mathrm{the}\mathrm{common}$$$$\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{three}\mathrm{circles}$$$$\left(3\right)\mathrm{if}c\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{this}\mathrm{we}\mathrm{get}\mathrm{a}‘‘{\mathrm{usual}}^{″}$$$$\mathrm{commoncircle}$$$$$$$$\mathrm{we}\mathrm{can}\mathrm{find}\mathrm{all}\mathrm{circles}\mathrm{touching}\mathrm{circles}\mathrm{a}\mathrm{\&}\mathrm{b}$$$$\mathrm{we}\mathrm{need}\mathrm{the}\mathrm{circle}\mathrm{d}\mathrm{with}\mathrm{exactly}\mathrm{one}\mathrm{common}$$$$\mathrm{point}\mathrm{with}\mathrm{circle}\mathrm{c}$$$$$$$$\left(1\right)\mathrm{circles}\mathrm{touching}\mathrm{circles}\mathrm{a}\mathrm{\&}\mathrm{b}\mathrm{from}\mathrm{outside}$$$$\mathrm{with}\mathrm{center}D=\left(\begin{array}{c}x\\ y\end{array}\right)$$$$\mid AD{\mid }^2={(a+d)}^2$$$$\mid BD{\mid }^2={(b+d)}^2$$$$$$$$\left(2\right)\mathrm{common}\mathrm{tangent}\mathrm{to}\mathrm{circles}\mathrm{a}\mathrm{\&}\mathrm{b}$$$$$$$$\left(3\right)\mathrm{circumcircles}\mathrm{of}\mathrm{circles}\mathrm{a}\mathrm{\&}\mathrm{b}$$$$\mathrm{with}\mathrm{center}D=\left(\begin{array}{c}x\\ y\end{array}\right)$$$$\mid AD{\mid }^2={(d-a)}^2$$$$\mid BD{\mid }^2={(d-b)}^2$$$$$$$$\mathrm{these}\mathrm{equations}\mathrm{are}\mathrm{getting}\mathrm{longer}\mathrm{from}\mathrm{step}$$$$\mathrm{to}\mathrm{step}\mathrm{but}{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{how}\mathrm{I}\mathrm{solved}\mathrm{it}$$

Commented by MJS last updated on 10/Sep/18

$$\mathrm{you}\mathrm{mean}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{technically}\mathrm{not}\mathrm{legible}\mathrm{or}{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{because}\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{post}\mathrm{the}\mathrm{works},\mathrm{only}\mathrm{the}$$$$\mathrm{results}?$$

Commented by ajfour last updated on 11/Sep/18

$$Centreofcircles:$$$$A(0,-R+a)$$$$B[(a+b)\cos \theta ,-R+a+(a+b)\sin \theta ]$$$$C[(a+c)\cos \varphi ,-R+a+(a+c)\sin \varphi ]$$$$$$$${(a+b)}^2\cos {}^2\theta +{[-R+a+(a+b)\sin \theta ]}^2$$$$={(R-b)}^2...\left(i\right)$$$${(a+c)}^2\cos {}^2\varphi +{[-R+a+(a+c)\sin \varphi ]}^2$$$$={(R-c)}^2...\left(ii\right)$$$${[(a+b)\cos \theta -(a+c)\cos \varphi ]}^2$$$$+{[(a+b)\sin \theta -(a+c)\sin \varphi ]}^2={(b+c)}^2$$$$....\left(iii\right)$$$$\Rightarrow {(a+b)}^2-2(R-a)(a+b)\sin \theta +{(R-a)}^2$$$$={(R-b)}^2$$$$\Rightarrow \sin \mathit{\theta }=\frac{{(\mathit{a}+\mathit{b})}^2+{(\mathit{R}-\mathit{a})}^2-{(\mathit{R}-\mathit{b})}^2}{2(\mathit{R}-\mathit{a})(\mathit{a}+\mathit{b})}$$$$\sin \mathit{\varphi }=\frac{{(\mathit{a}+\mathit{c})}^2+{(\mathit{R}-\mathit{a})}^2-{(\mathit{R}-\mathit{c})}^2}{2(\mathit{R}-\mathit{a})(\mathit{a}+\mathit{c})}$$$$\Rightarrow \sin \theta =\frac{a(a+b)-R(a-b)}{R(a+b)-a(a+b)}$$$$\Rightarrow \sin \varphi =\frac{a(a+c)-R(a-c)}{R(a+c)-a(a+c)}$$$$\Rightarrow \sin \theta =\frac{\frac{a}{R}-\frac{a-b}{a+b}}{1-\frac{a}{R}}=\frac{x-p}{1-x}$$$$\sin \varphi =\frac{\frac{a}{R}-\frac{a-c}{a+c}}{1-\frac{a}{R}}=\frac{x-q}{1-x}$$$$\mathrm{\&}$$$$\mathit{E}\mathit{q}.\left(\mathit{i}\mathit{i}\mathit{i}\right)again$$$${(\mathit{a}+\mathit{b})}^2+{(\mathit{a}+\mathit{c})}^2-{(\mathit{b}+\mathit{c})}^2$$$$=2(\mathit{a}+\mathit{b})(\mathit{a}+\mathit{c})\cos (\mathit{\theta }-\mathit{\varphi })$$$$\Rightarrow {(b-c)}^2-{(b+c)}^2$$$$=-2(a+b)(a+c)[1-\cos (\theta -\varphi )]$$$$\Rightarrow 1-\cos (\theta -\varphi )=\frac{2bc}{(a+b)(a+c)}$$$$.....$$

Commented by ajfour last updated on 10/Sep/18

$$willneedtodownloadandtake$$$$printouts\left(asihaveasmallscreenphone\right).$$

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