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Circles

DrunkenOldFool

New member
Feb 6, 2012
20
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 $x-y=7$.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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. :)
 

sbhatnagar

Active member
Jan 27, 2012
95
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:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
With problems like this, it's always good to draw a diagram (click on the diagram to embiggen it).

[graph]psuayl2bwh[/graph]

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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725

DrunkenOldFool

New member
Feb 6, 2012
20
With problems like this, it's always good to draw a diagram (click on the diagram to embiggen it).

[graph]psuayl2bwh[/graph]

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.
Thank You! This was extremely helpful.