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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
$eqn big circle x^2 +y^2 =R^2 and area πR^2 small circle centre(−a,0) radius r (x+a)^2 +y^2 =r^2 small circle pass through{ (R−18),0},{0,(R−10)}and {0,−(R−10)}(−R,0} (R−18+a)^2 +0^2 =r^2 (0+a)^2 +(R−10)^2 =r^2 (0+a)^2 +{−(R−10)}^2 =r^2 (−R+a)^2 +0^2 =r^2 a^2 +(R−10)^2 =r^2 (R−18+a)^2 =a^2 +(R−10)^2 (50−18+a)^2 =a^2 +(50−10)^2 32^2 +64a+a^2 =a^2 +1600 64a=1600−32^2 =40^2 −32^2 =72×8 a=((72×8)/(64))=9 R×(R−18)=(R−10)(R−10)⇚look here for R R^2 −18R=R^2 −20R+100 2R=100 R=50 a^2 +(R−10)^2 =r^2 81+1600=r^2 r^2 =1681 r=41 required green area=π(R^2 −r^2 ) =π(50^2 −41^2 )=π×91×9=819π$
$e q n b i g c i r c l e x^{2} + y^{2} = R^{2} a n d a r e a π R^{2}$ $s m a l l c i r c l e c e n t r e (- a, 0) r a d i u s r$ ${(x + a)}^{2} + y^{2} = r^{2}$ $s m a l l c i r c l e p a s s t h r o u g h {(R - 18), 0}, {0, (R - 10)} a n d {0, - (R - 10)} (- R, 0}$ ${(R - 18 + a)}^{2} + 0^{2} = r^{2}$ ${(0 + a)}^{2} + {(R - 10)}^{2} = r^{2}$ ${(0 + a)}^{2} + {- (R - 10)}^{2} = r^{2}$ ${(- R + a)}^{2} + 0^{2} = r^{2}$ $a^{2} + {(R - 10)}^{2} = r^{2}$ ${(R - 18 + a)}^{2} = a^{2} + {(R - 10)}^{2}$ ${(50 - 18 + a)}^{2} = a^{2} + {(50 - 10)}^{2}$ $32^{2} + 64 a + a^{2} = a^{2} + 1600$ $64 a = 1600 - 32^{2} = 40^{2} - 32^{2} = 72 \times 8$ $a = \frac{72 \times 8}{64} = 9$ $R \times (R - 18) = (R - 10) (R - 10) ⇚ l o o k h e r e f o r R$ $R^{2} - 18 R = R^{2} - 20 R + 100$ $2 R = 100 R = 50$ $a^{2} + {(R - 10)}^{2} = r^{2}$ $81 + 1600 = r^{2}$ $r^{2} = 1681 r = 41$ $r e q u i r e d g r e e n a r e a = π (R^{2} - r^{2})$ $= π (50^{2} - 41^{2}) = π \times 91 \times 9 = 819 π$
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