Thank you geno!
AX · XB = XY · XY
This is the secant-tangent theorem. But is that really true?
I'll do the math:
The radius of the circle should be 6, then:
\(f(x) =\pm \sqrt{6^2-(x-8)^2}-6\)
\(g(x) = -\sqrt{15^2-x^2}\)
Where do the circles intersect?
\(f(x)=g(x)\\ \pm \sqrt{6^2-(x-8)^2}-6= -\sqrt{15^2-x^2}\\ I)\ 6^2-(x-8)^2+12\times+\sqrt{6^2-(x-8)^2}+36=15^2-x^2\\ II)\ 6^2-(x-8)^2+12\times-\sqrt{6^2-(x-8)^2}+36=15^2-x^2\\ \)
\(Wolfram\ Alpha\ did\ the\ math\ for\ me:\\ I)\\ x_1=\frac{289}{25}-\frac{3\sqrt{6479}}{100}\\ x_1=9.145233\\ II)\\ x_2=\frac{289}{25}+\frac{3\sqrt{6479}}{100}\\ \color{blue}x_2=13.974767 \)
\(Only\ one\ Point\ of\ intersection\ is\ needed\ for\ the\ "proof".\\ I\ take\ x_2. \)
\(g(x) =y_2= -\sqrt{15^2-x_2\ ^2}\\ \color{blue}y_2=-5.45031\)
\(The\ secant\ corresponds\ to\ the\ functional\ equation:\\ \color{blue}f_{sec}(x)=-\frac{y_2}{x_2}\cdot x=-0.39001\cdot x\\ The\ length\ of\ the\ secant\ is\\ l_{sec }=\sqrt{13.975^2-5.45)^2}\\ \color{blue}lsec =15.000 \)
Where does the secant f(x) intersect?
\( \sqrt{6^2-(x-8)^2}-6=-0.39001x\\ \sqrt{6^2-(x-8)^2}=-0.39001x+6\\ 6^2-(x-8)^2=0,152x^2-4.68x+36\\ 6^2-x^2+16x-64=0,152x^2-4.68x+36\\ 1.152x^2-20.68x+64=0\\ \color{blue}x_{1,2}\in \{3.975,13.976\}\\ \color{blue}y_{1,2}\in \{-1.543,-5.450\}\)
\(\overline{AX}=15\\ \overline{BX}=\sqrt{3.975^2+1.534^2}=4.267\\ \color{blue} \overline{BX} =4.267\\ \color{blue}\overline{XY}=8\\\)
\(claim:\\ \color{blue}\overline{AX}\times \overline{BX}=\overline{XY}^2\\ confirmation: \\ 15\times4.267=8^2\\ \color{blue}64=64\)
quod erat demonstrandum
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