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Question Number 71698 by TawaTawa last updated on 18/Oct/19

Commented by MJS last updated on 19/Oct/19

$$\mathrm{we}\mathrm{had}\mathrm{a}\mathrm{similar}\mathrm{example}\mathrm{a}\mathrm{few}\mathrm{weeks}\mathrm{ago}$$

Commented by TawaTawa last updated on 19/Oct/19

$$\mathrm{Help}\mathrm{sir}$$

Answered by MJS last updated on 19/Oct/19

$$\mathrm{general}\mathrm{solution}$$$$\mathrm{let}\mathrm{the}\mathrm{right}\mathrm{handed}\mathrm{difference}\mathrm{be}p\mathrm{and}\mathrm{the}$$$$\mathrm{difference}\mathrm{on}\mathrm{bottom}\mathrm{and}\mathrm{top}\mathrm{be}q$$$$\Rightarrow$$$$\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{big}\mathrm{circle}C\mathrm{is}R$$$$\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{small}\mathrm{circle}c\mathrm{is}r=R-\frac{p}{2}$$$$C:x^2+y^2-R^2=0$$$$c:{(x-\frac{p}{2})}^2+y^2-{(R-\frac{p}{2})}^2=0\iff$$$$\iff x^2+y^2-px-R^2+pR=0$$$$\mathrm{we}\mathrm{know}\mathrm{that}\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}0\\ R-q\end{array}\right)\in c$$$$\mathrm{put}x=0\wedge y=R-q\mathrm{into}c\Rightarrow$$$$(p-2q)R+q^2=0\Rightarrow R=\frac{q^2}{2q-p}\Rightarrow r=\frac{{(p-q)}^2+q^2}{2(2q-p)}$$$$\Rightarrow \mathrm{green}\mathrm{area}=\mathrm{area}\left(C\right)-\mathrm{area}\left(c\right)=$$$$=\frac{p({(p-q)}^2+3q^2)}{4(2q-p)}\pi$$$$$$$$\mathrm{in}\mathrm{our}\mathrm{case}$$$$p=18q=10$$$$R=50r=41$$$$\mathrm{green}\mathrm{area}=819\pi$$

Commented by TawaTawa last updated on 19/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa last updated on 21/Oct/19

$$\mathrm{Sir},\mathrm{help}\mathrm{me}\mathrm{check}\mathrm{question}71837$$

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