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Question Number 90499 by ajfour last updated on 24/Apr/20

Commented by ajfour last updated on 24/Apr/20

$$★Similartoapreviousquestion.$$$$Ifparabolaisy=x^2,and△ABC$$$$isequilateral,findeq.ofcircle.$$

Commented by mr W last updated on 24/Apr/20

$$foreverysidelengthswegetacircle,$$$$sonouniquesolution.$$

Commented by ajfour last updated on 24/Apr/20

$$thenwecantrychoosingthe$$$$circlethatpassesthroughthe$$$$origin,Oeven.$$$$Sir,doyouknowtheQ.noofthe$$$$previoussimilarquestion?$$

Commented by mr W last updated on 24/Apr/20

$$89281,62938$$

Commented by ajfour last updated on 24/Apr/20

$$thanks,sir.$$

Commented by mr W last updated on 25/Apr/20

$$yoursolutionis62938isamilestone$$$$forsolvingallthesequestions!$$

Answered by mr W last updated on 24/Apr/20

$$s=sidelengthofABC$$$$(h,k)=centerofABC$$$$h=\sqrt{\frac{s^2}{48}-\frac{1}{4}}$$$$k=\frac{3s^2}{16}-\frac{1}{4}$$$$x_i=2\sqrt{-p}\sin (\frac{1}{3}{\sin }^{-1}\frac{q}{\sqrt{-p^3}}+\frac{2i\pi }{3})+h$$$$i=0,1,2forC,A,B$$$$with$$$$p=\frac{h^2-k}{2}$$$$q=\frac{h(5h^2-3k)}{4}+\frac{3h^2+3k^2-s^2}{18h}$$$$x_C=2\sqrt{-p}\sin \left(\frac{1}{3}{\sin }^{-1}\frac{q}{\sqrt{-p^3}}\right)+h$$$$x_C^2(1+x_C^2)=s^2$$$$\Rightarrow s=8.0429$$

Commented by mr W last updated on 24/Apr/20

Commented by ajfour last updated on 25/Apr/20

$$\mathcal{T}hanks\mathcal{S}ir.$$

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