Find the equations of the sides of square inscribed in the circle $3(x^2+y^2)=4$, one of whose sides is parallel to the line $x-y=7$.
Find the equations of the sides of square inscribed in the circle $3(x^2+y^2)=4$, one of whose sides is parallel to the line $x-y=7$.
The circle has radius $\frac{2}{\sqrt{3}}$. Since one of the sides is parallel to $x-y=7$, another side is parallel as well and the two left are perpendicular. This gives you a clue about all four equations you need to find.
Let one of the side be $y=x+k$ (Note that the line $y=x+k$ is always paralell to $y=x-7$) . The half of the length of the side of square will be equal to the perpendicular distance from the origin to the line $x+k=y$.
\[\frac{l}{2}= \Bigg| \frac{0-0+k}{\sqrt{2}}\Bigg|=\frac{|k|}{\sqrt{2}}\]
The length of each side is $l=\sqrt{2}|k|$ and the diameter of the circle is $\frac{4}{\sqrt{3}}$. Also, the diagonal of a square is $\sqrt{2}$ times its length of side. Therefore
\[\sqrt{2} \times \sqrt{2}|k|=\frac{4}{\sqrt{3}}\]
From here we get two values of $k$ i.e $2/\sqrt{3},-2/\sqrt{3}$. So two sides are $y=x+\frac{2}{\sqrt{3}},y=x-\frac{2}{\sqrt{3}}$.
Last edited by sbhatnagar; November 5th, 2012 at 12:13.
With problems like this, it's always good to draw a diagram (click on the diagram to embiggen it).
The line $x-y=7$ has gradient 1, so the sides of the square will have gradient 1 and $-1.$ The circle is centred at the origin and has radius $2/\sqrt3$, so the vertices of the square will be at the points $(\pm2/\sqrt3,0)$ and $(0,\pm2/\sqrt3).$ You want the equations of the lines with gradient $\pm1$ going through those points, and with that information you can write down the answers.