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Question-120790

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Question Number 120790 by peter frank last updated on 02/Nov/20
Commented by TANMAY PANACEA last updated on 02/Nov/20
eqn big circle x^2 +y^2 =R^2 and area πR^2 small circle centre(−a,0) radius r (x+a)^2 +y^2 =r^2 small circle pass through{ (R−18),0},{0,(R−10)}and {0,−(R−10)}(−R,0} (R−18+a)^2 +0^2 =r^2 (0+a)^2 +(R−10)^2 =r^2 (0+a)^2 +{−(R−10)}^2 =r^2 (−R+a)^2 +0^2 =r^2 a^2 +(R−10)^2 =r^2 (R−18+a)^2 =a^2 +(R−10)^2 (50−18+a)^2 =a^2 +(50−10)^2 32^2 +64a+a^2 =a^2 +1600 64a=1600−32^2 =40^2 −32^2 =72×8 a=((72×8)/(64))=9 R×(R−18)=(R−10)(R−10)⇚look here for R R^2 −18R=R^2 −20R+100 2R=100 R=50 a^2 +(R−10)^2 =r^2 81+1600=r^2 r^2 =1681 r=41 required green area=π(R^2 −r^2 ) =π(50^2 −41^2 )=π×91×9=819π
$${eqn}\:{big}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:{and}\:{area}\:\pi{R}^{\mathrm{2}} \\ $$$${small}\:{circle}\:{centre}\left(−{a},\mathrm{0}\right)\:{radius}\:{r} \\ $$$$\left({x}+{a}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${small}\:{circle}\:{pass}\:{through}\left\{\:\left({R}−\mathrm{18}\right),\mathrm{0}\right\},\left\{\mathrm{0},\left({R}−\mathrm{10}\right)\right\}{and}\:\left\{\mathrm{0},−\left({R}−\mathrm{10}\right)\right\}\left(−{R},\mathrm{0}\right\} \\ $$$$\left({R}−\mathrm{18}+{a}\right)^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(\mathrm{0}+{a}\right)^{\mathrm{2}} +\left({R}−\mathrm{10}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(\mathrm{0}+{a}\right)^{\mathrm{2}} +\left\{−\left({R}−\mathrm{10}\right)\right\}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(−{R}+{a}\right)^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\left({R}−\mathrm{10}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left({R}−\mathrm{18}+{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({R}−\mathrm{10}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{50}−\mathrm{18}+{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{50}−\mathrm{10}\right)^{\mathrm{2}} \\ $$$$\mathrm{32}^{\mathrm{2}} +\mathrm{64}{a}+{a}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{1600} \\ $$$$\mathrm{64}{a}=\mathrm{1600}−\mathrm{32}^{\mathrm{2}} =\mathrm{40}^{\mathrm{2}} −\mathrm{32}^{\mathrm{2}} =\mathrm{72}×\mathrm{8} \\ $$$${a}=\frac{\mathrm{72}×\mathrm{8}}{\mathrm{64}}=\mathrm{9} \\ $$$$ \\ $$$${R}×\left({R}−\mathrm{18}\right)=\left({R}−\mathrm{10}\right)\left({R}−\mathrm{10}\right)\Lleftarrow{look}\:{here}\:{for}\:{R} \\ $$$${R}^{\mathrm{2}} −\mathrm{18}{R}={R}^{\mathrm{2}} −\mathrm{20}{R}+\mathrm{100} \\ $$$$\mathrm{2}{R}=\mathrm{100}\:\:\:{R}=\mathrm{50} \\ $$$${a}^{\mathrm{2}} +\left({R}−\mathrm{10}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{81}+\mathrm{1600}={r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} =\mathrm{1681}\:\:{r}=\mathrm{41} \\ $$$${required}\:{green}\:{area}=\pi\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$$=\pi\left(\mathrm{50}^{\mathrm{2}} −\mathrm{41}^{\mathrm{2}} \right)=\pi×\mathrm{91}×\mathrm{9}=\mathrm{819}\pi \\ $$

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