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Question Number 65212 by ajfour last updated on 26/Jul/19

Commented by ajfour last updated on 26/Jul/19

$$ForPtobeunique,determine$$$$howa,b,carerelated.$$

Answered by ajfour last updated on 26/Jul/19

$$letB(0,0)$$$$C(a,0),D(\frac{ac}{b+c},0)$$$$\underset{-}{ForA(h,k)}$$$$x^2+y^2=c^2$$$${(x-a)}^2+y^2=b^2$$$$\Rightarrow a(2x-a)=c^2-b^2$$$$\Rightarrow x_A=h=\frac{a^2-b^2+c^2}{2a}$$$$y_A=k=\sqrt{c^2-{\left(\frac{a^2-b^2+c^2}{2a}\right)}^2}$$$$\underset{-}{eq.ofBE}y=\frac{kx}{a+h}$$$$\underset{-}{eq.ofCF}y=-\frac{h}{k}(x-a)$$$$\underset{-}{eq.ofAD}y=\frac{k(x-\frac{ac}{b+c})}{(h-\frac{ac}{b+c})}$$$$ForintersectionofBE,CF$$$$\frac{kx}{a+h}=-\frac{h}{k}(x-a)$$$$\Rightarrow k^{2}x+h(a+h)x=ah(a+h)$$$$\Rightarrow x_P=\frac{ah(a+h)}{k^2+h(a+h)}$$$$y_P=\frac{ahk}{k^2+h(a+h)}$$$$IfPisunique,PevenliesonAD$$$$\Rightarrow \frac{ahk}{k^2+h(a+h)}=\frac{k(\frac{ah(a+h)}{k^2+h(a+h)}-\frac{ac}{b+c})}{(h-\frac{ac}{b+c})}$$$$whereh=\frac{a^2-b^2+c^2}{2a}$$$$andk=\sqrt{c^2-{\left(\frac{a^2-b^2+c^2}{2a}\right)}^2}$$$$withlittlerearrangements$$$$aftersubstitutionandif$$$$k\ne 0,a\ne 0,iarrivedat$$$$ah(b+c)=c(h^2+k^2)$$$$\Rightarrow a(b+c)\left(\frac{a^2-b^2+c^2}{2a}\right)=c^3$$$$\Rightarrow (b+c)(a^2-b^2+c^2)=2c^3$$$$\Rightarrow (b+c)(a^2-b^2)=c^2(c-b)$$$$\Rightarrow \frac{a^2-b^2}{c^2}=\frac{c-b}{c+b}.$$

Commented by mr W last updated on 26/Jul/19

$$nicesolution!$$

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