Question-153857

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×OB=ab AB=b−a M=midpoint of AB OM=((a+b)/2) KM^2 =R^2 −(((AB)/2))^2 =R^2 −(((b−a)/2))^2 OK^2 =OD^2 +R^2 =ab+R^2 tan θ=(((R/( (√(ab))))+((√(R^2 −(((b−a)/2))^2 ))/((a+b)/2)))/(1−(R/( (√(ab))))×((√(R^2 −(((b−a)/2))^2 ))/((a+b)/2)))) ⇒(((a+b)R+2(√(ab))(√(R^2 −(((b−a)/2))^2 )))/((a+b)(√(ab))−2R(√(R^2 −(((b−a)/2))^2 ))))=tan θ we get R from this equation. k_1 =OK_1 =(R/(sin θ)) k_2 =OK_2 =(√(ab))−(R/(tan θ))

$${OD}^{\mathrm{2}} ={OA}×{OB}={ab} \\ $$$${AB}={b}−{a} \\ $$$${M}={midpoint}\:{of}\:{AB} \\ $$$${OM}=\frac{{a}+{b}}{\mathrm{2}} \\ $$$${KM}^{\mathrm{2}} ={R}^{\mathrm{2}} −\left(\frac{{AB}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} −\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${OK}^{\mathrm{2}} ={OD}^{\mathrm{2}} +{R}^{\mathrm{2}} ={ab}+{R}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=\frac{\frac{{R}}{\:\sqrt{{ab}}}+\frac{\sqrt{{R}^{\mathrm{2}} −\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\frac{{a}+{b}}{\mathrm{2}}}}{\mathrm{1}−\frac{{R}}{\:\sqrt{{ab}}}×\frac{\sqrt{{R}^{\mathrm{2}} −\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\frac{{a}+{b}}{\mathrm{2}}}} \\ $$$$\Rightarrow\frac{\left({a}+{b}\right){R}+\mathrm{2}\sqrt{{ab}}\sqrt{{R}^{\mathrm{2}} −\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\left({a}+{b}\right)\sqrt{{ab}}−\mathrm{2}{R}\sqrt{{R}^{\mathrm{2}} −\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}=\mathrm{tan}\:\theta \\ $$$${we}\:{get}\:{R}\:{from}\:{this}\:{equation}. \\ $$$${k}_{\mathrm{1}} ={OK}_{\mathrm{1}} =\frac{{R}}{\mathrm{sin}\:\theta} \\ $$$${k}_{\mathrm{2}} ={OK}_{\mathrm{2}} =\sqrt{{ab}}−\frac{{R}}{\mathrm{tan}\:\theta} \\ $$

Commented by mathdanisur last updated on 11/Sep/21

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

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