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Question Number 153857 by mathdanisur last updated on 11/Sep/21

Commented by Tawa11 last updated on 11/Sep/21

$$\mathrm{Weldone}\mathrm{sir}.$$

Answered by mr W last updated on 11/Sep/21

$${OD}^2=OA\times OB=ab$$$$AB=b-a$$$$M=midpointofAB$$$$OM=\frac{a+b}{2}$$$${KM}^2=R^2-{\left(\frac{AB}{2}\right)}^2=R^2-{\left(\frac{b-a}{2}\right)}^2$$$${OK}^2={OD}^2+R^2=ab+R^2$$$$\tan \theta =\frac{\frac{R}{\sqrt{ab}}+\frac{\sqrt{R^2-{\left(\frac{b-a}{2}\right)}^2}}{\frac{a+b}{2}}}{1-\frac{R}{\sqrt{ab}}\times \frac{\sqrt{R^2-{\left(\frac{b-a}{2}\right)}^2}}{\frac{a+b}{2}}}$$$$\Rightarrow \frac{(a+b)R+2\sqrt{ab}\sqrt{R^2-{\left(\frac{b-a}{2}\right)}^2}}{(a+b)\sqrt{ab}-2R\sqrt{R^2-{\left(\frac{b-a}{2}\right)}^2}}=\tan \theta$$$$wegetRfromthisequation.$$$$k_1={OK}_1=\frac{R}{\sin \theta }$$$$k_2={OK}_2=\sqrt{ab}-\frac{R}{\tan \theta }$$

Commented by mathdanisur last updated on 11/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{thank}\mathrm{you}\mathrm{Ser}$$

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