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Question Number 75291 by 21042004 last updated on 09/Dec/19

Answered by mind is power last updated on 09/Dec/19

$$\mathrm{let}\mathrm{A}(\mathrm{a},0),\mathrm{B}=(\mathrm{b},0)$$$$\mathrm{Equation}\mathrm{of}\mathrm{circl}\mathrm{center}\mathrm{in}\mathrm{A}$$$$\mathrm{condition}3\mathrm{AB}<2\mathrm{R}$$$$\mathrm{if}\mathrm{not}{\mathrm{C}}_{\mathrm{A}}\cap {\mathrm{C}}_{\mathrm{B}}=\varnothing$$$${(\mathrm{x}-\mathrm{a})}^2+{\mathrm{y}}^2={\mathrm{R}}^2....{\mathrm{C}}_{\mathrm{A}}$$$${(\mathrm{x}-\mathrm{b})}^2+{\mathrm{y}}^2={\mathrm{R}}^2...{\mathrm{C}}_{\mathrm{B}}$$$${\mathrm{C}}_{\mathrm{A}}\cap {\mathrm{C}}_{\mathrm{B}}\iff {(\mathrm{x}-\mathrm{b})}^2-{(\mathrm{x}-\mathrm{a})}^2=0\Rightarrow \mathrm{x}=\frac{\mathrm{a}+\mathrm{b}}{2}$$$${\mathrm{y}}^2={\mathrm{R}}^2-{\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)}^2\Rightarrow \mathrm{y}=\underset{-}{+}\sqrt{{\mathrm{R}}^2-{\left(\frac{\mathrm{a}-\mathrm{b}}{2}\right)}^2}=\underset{-}{+}\sqrt{\alpha }$$$${\mathrm{C}}_{\mathrm{A}}\cap {\mathrm{C}}_{\mathrm{B}}=\{\mathrm{S},\mathrm{D}\}\mathrm{S}(\frac{\mathrm{a}+\mathrm{b}}{2},\sqrt{}\alpha ),\mathrm{D}(\frac{\mathrm{a}+\mathrm{b}}{2},-\sqrt{}\alpha )$$$$\mathrm{O}=(\frac{\mathrm{a}+\mathrm{b}}{2},0)$$$$\mid \mathrm{OA}\mid =\mid \frac{\mathrm{a}-\mathrm{b}}{2}\mid =\mid \mathrm{OB}\mid =\mid \frac{\mathrm{b}-\mathrm{a}}{2}\mid$$$${\mathrm{OA}}^2+{\mathrm{OS}}^2=\alpha +={\mathrm{R}}^2={\mathrm{AS}}^2\Rightarrow \mathrm{AOS}\mathrm{is}\mathrm{rectangle}\mathrm{in}\mathrm{O}\mathrm{pythagor}\mathrm{theorem}$$$$$$$$$$$$$$$$$$

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