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Question Number 98098 by bobhans last updated on 11/Jun/20

$$\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{chord}\mathrm{cut}$$$$\mathrm{off}\mathrm{by}\mathrm{y}=2\mathrm{x}+1\mathrm{from}\mathrm{circle}{\mathrm{x}}^2+{\mathrm{y}}^2=2$$

Answered by john santu last updated on 11/Jun/20

Commented by john santu last updated on 11/Jun/20

$$\stackrel{⌢}{\mathrm{A}}\stackrel{⌢}{\mathrm{B}}=\underset{-1}{\stackrel{0.2}{\int }}\sqrt{1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}\mathrm{dx}$$$$\iff \frac{\mathrm{d}}{\mathrm{dx}}[{\mathrm{x}}^2+{\mathrm{y}}^2=2]$$$$2\mathrm{x}+2\mathrm{yy}{}^{\prime }=0;{\mathrm{y}}^{\prime }=-\frac{\mathrm{x}}{\mathrm{y}}$$$${\left({\mathrm{y}}^{\prime }\right)}^2=\frac{{\mathrm{x}}^2}{2-{\mathrm{x}}^2}.$$$$\stackrel{⌢}{\mathrm{A}}\stackrel{⌢}{\mathrm{B}}=\underset{-1}{\stackrel{0.2}{\int }}\sqrt{1+\frac{{\mathrm{x}}^2}{2-{\mathrm{x}}^2}}\mathrm{dx}$$$$=\underset{-1}{\stackrel{0.2}{\int }}\sqrt{\frac{2}{2-{\mathrm{x}}^2}}\mathrm{dx}=\sqrt{2}\underset{-1}{\stackrel{0.2}{\int }}\frac{\mathrm{dx}}{\sqrt{2-{\mathrm{x}}^2}}$$$$\mathrm{set}\mathrm{x}=\sqrt{2}\sin \mathrm{t}$$$$[\int \frac{\sqrt{2}\cos \mathrm{t}\mathrm{dt}}{\sqrt{2(1-\sin {}^2\mathrm{t})}}={\sin }^{-1}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)]$$$$\mathrm{then}\stackrel{⌢}{\mathrm{A}}\stackrel{⌢}{\mathrm{B}}=\sqrt{2}{\left[{\sin }^{-1}\left(\frac{\mathrm{x}}{\sqrt{2}}\right)\right]}_{-1}^{0.2}$$$$=\sqrt{2}({\sin }^{-1}\left(\frac{1}{5\sqrt{2}}\right)+\frac{\pi }{4})$$

Answered by mr W last updated on 11/Jun/20

$$y=2x+1$$$$x^2+{(2x+1)}^2=2$$$$5x^2+4x-1=0$$$$x_1+x_2=-\frac{4}{5}$$$$x_1{x}_2=-\frac{1}{5}$$$${(x_2-x_1)}^2={(x_1+x_2)}^2-4x_1{x}_2=\frac{36}{25}$$$$\mathrm{\Delta }x=x_2-x_1=\frac{6}{5}$$$$L_{chord}=\sqrt{2^2+1^2}\mathrm{\Delta }x=\frac{6\sqrt{5}}{5}$$

Commented by bobhans last updated on 11/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Answered by 1549442205 last updated on 11/Jun/20

$$\mathrm{Intersection}\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{y}=2\mathrm{x}+1\mathrm{and}$$$$\mathrm{the}\mathrm{circle}{\mathrm{x}}^2+{\mathrm{y}}^2=2\mathrm{have}\mathrm{the}\mathrm{cordinates}$$$$\mathrm{be}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}:$$$$\{\begin{array}{l}\mathrm{y}=2\mathrm{x}+1\left(1\right)\\ {\mathrm{x}}^2+{\mathrm{y}}^2=2\left(2\right)\end{array}$$$$\mathrm{Replace}\left(1\right)\mathrm{into}\left(2\right)\mathrm{we}\mathrm{get}{\mathrm{x}}^2+{(2\mathrm{x}+1)}^2=2$$$$\iff {5\mathrm{x}}^2+4\mathrm{x}-1=0\iff {\mathrm{x}}_{\mathrm{A}}=-1,{\mathrm{x}}_{\mathrm{B}}=\frac{1}{5}\Rightarrow {\mathrm{y}}_{\mathrm{A}}=-1,{\mathrm{y}}_{\mathrm{B}}=\frac{7}{5}.\mathrm{We}\mathrm{get}$$$$\mathrm{A}(-1;-1),\mathrm{B}(\frac{1}{5};\frac{7}{5})$$$$\mathrm{AB}=\sqrt{{({\mathrm{x}}_{\mathrm{A}}-{\mathrm{x}}_{\mathrm{B}})}^2+{({\mathrm{y}}_{\mathrm{A}}-{\mathrm{y}}_{\mathrm{B}})}^2}$$$$=\sqrt{{\left(\frac{6}{5}\right)}^2+{\left(\frac{12}{5}\right)}^2}=\sqrt{\frac{36+144}{25}}=\sqrt{\frac{180}{25}}=\frac{6\sqrt{5}}{5}$$$$\mathrm{Thus},\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{chord}\mathrm{AB}\mathrm{is}\frac{6\sqrt{5}}{5}$$

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