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Question Number 65022 by ajfour last updated on 24/Jul/19

Commented by MJS last updated on 24/Jul/19

$$\mathrm{I}\mathrm{used}\mathrm{a}\mathrm{calculator}.{\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{got}\mathrm{an}\mathrm{old}\mathrm{TI}-89$$$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{any}\mathrm{calculator}\mathrm{which}\mathrm{is}\mathrm{able}\mathrm{to}$$$$\mathrm{approximately}\mathrm{solve}\mathrm{equations}\mathrm{and}\mathrm{integrals}$$$$\mathrm{select}\mathrm{a}\mathrm{value}\mathrm{for}r$$$$\mathrm{calculate}P,Q\mathrm{and}\mathrm{the}\mathrm{integrals}$$$$\mathrm{it}\mathrm{takes}\mathrm{some}\mathrm{time}\mathrm{though}...$$

Commented by ajfour last updated on 07/Aug/19

$$AreaofregionA=AreaofregionB.$$$$Findradiusofcircle.$$

Commented by ajfour last updated on 24/Jul/19

$$YouareinvincibleSir,buthow$$$$couldyouevenapproximately$$$$arriveatavalueof\mathit{r};doexplain$$$$Sir.$$

Commented by MJS last updated on 24/Jul/19

$$\mathrm{I}\mathrm{can}\mathrm{only}\mathrm{solve}\mathrm{approximately}\mathrm{and}\mathrm{get}$$$$r\approx .9269083316$$

Answered by ajfour last updated on 07/Aug/19

$${(x-r)}^2+{(y-r)}^2=r^2$$$$x^2+y^2-2r(x+y)+r^2=0$$$$butalsoy=x^2$$$$\Rightarrow x^2+x^4-2r(x+x^2)+r^2=0$$$$x^4+(1-2r)x^2-2rx+r^2=0$$$$a=1-2r,b=-2r,c=r^2$$$$p^6+2{ap}^4+(a^2-4c)p^2-b^2=0$$$$\Rightarrow p^6+2(1-2r)p^4+(1-4r)p^2-4r^2=0$$$$let{p}^2=z-\frac{2}{3}(1-2r)$$$$z^3+\frac{4}{3}{(1-2r)}^{2}z-\frac{8}{27}{(1-2r)}^3$$$$+2(1-2r)[-\frac{4}{3}(1-2r)z+\frac{4}{9}{(1-2r)}^2]$$$$+(1-4r)[z-\frac{2}{3}(1-2r)]-4r^2=0$$$$\Rightarrow z^3-z[\frac{4}{3}{(1-2r)}^2-(1-4r)]$$$$+{(1-2r)}^3[\frac{8}{9}-\frac{8}{27}]$$$$-\frac{2}{3}(1-4r)(1-2r)-4r^2=0$$$$....$$

Commented by ajfour last updated on 07/Aug/19

$$hopeless..tocarry$$

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