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Question-137374




Question Number 137374 by oooooooo last updated on 02/Apr/21
Answered by mr W last updated on 02/Apr/21
let OA=OD=R=radius  ((R/( (√2)))−4)^2 +((R/( (√2)))+3)^2 =R^2    ...(i)  ((R/( (√2)))−x)^2 +((R/( (√2)))+5)^2 =R^2    ...(ii)  (i):  −(√2)R+25=0 ⇒R=((25)/( (√2)))  (ii):  (((25)/2)−x)^2 +(((25)/2)+5)^2 =((25^2 )/2)  (((25)/2)−x)^2 =((5/2))^2   ((25)/2)−x=±(5/2)  x=((25)/2)±(5/2)=15 or 10
$${let}\:{OA}={OD}={R}={radius} \\ $$$$\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−\mathrm{4}\right)^{\mathrm{2}} +\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}+\mathrm{3}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}−{x}\right)^{\mathrm{2}} +\left(\frac{{R}}{\:\sqrt{\mathrm{2}}}+\mathrm{5}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right): \\ $$$$−\sqrt{\mathrm{2}}{R}+\mathrm{25}=\mathrm{0}\:\Rightarrow{R}=\frac{\mathrm{25}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left({ii}\right): \\ $$$$\left(\frac{\mathrm{25}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} +\left(\frac{\mathrm{25}}{\mathrm{2}}+\mathrm{5}\right)^{\mathrm{2}} =\frac{\mathrm{25}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{25}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} =\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{25}}{\mathrm{2}}−{x}=\pm\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{25}}{\mathrm{2}}\pm\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{15}\:{or}\:\mathrm{10} \\ $$

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