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Sagot :
Given
[tex]y=-\frac{x}{4}[/tex]We know that the general equation for a circle is given by:
[tex](x-h)^2+(x-k)^2=r^2[/tex]Since the circle for this case is centered at the origin then h,k =0 and we have this:
[tex]x^2+y^2=2^2[/tex]For this case we can replace the formula for y from the line into equation (1) and we got:
[tex]\begin{gathered} x^2+(-\frac{x}{4})^2=4 \\ \\ x^2+\frac{x^2}{16}=4 \\ multiply\text{ all through by 16} \\ 16x^2+x^2=64 \\ 17x^2=64 \\ \end{gathered}[/tex]Divide both sides by 17
[tex]\begin{gathered} \frac{17x^2}{17}=\frac{64}{17} \\ \\ x^2=\frac{64}{17} \\ \\ \\ Take\text{ the square root} \\ x=\sqrt{\frac{64}{17}} \\ \\ x=\frac{8}{\sqrt{17}} \\ Rationalize \\ \\ x=\pm\frac{8\sqrt{17}}{17} \end{gathered}[/tex]Now we just need to replace into the original equation for the line and we get the y coordinates like this:
[tex]\begin{gathered} y=-\frac{(\frac{8\sqrt{17}}{17})}{4} \\ \\ y=-\frac{2\sqrt{17}}{17} \end{gathered}[/tex]and
[tex]\begin{gathered} y=-\frac{(\frac{-8\sqrt{17}}{17})}{17} \\ \\ y=\frac{8\sqrt{17}}{289} \end{gathered}[/tex]The final answer
[tex](\frac{8\sqrt{17}}{17},-\frac{2\sqrt{17}}{17}),(\frac{8\sqrt{17}}{17},\frac{8\sqrt{17}}{289})[/tex]
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