coordinate geometry

Feb 2010
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the circle x^2+y^2=a^2 cuts off intercept on the straight line lx+my=1 which subtends an angle 45 at the origin.show that a^2(l^2+m^2)=4-2*1.414
 
Jun 2007
1,252
683
Medgidia, Romania
Let \(\displaystyle A(x_1,y_1), \ B(x_1,y_1)\) be the intersect points between the line and the circle. Then \(\displaystyle \widehat{AOB}=45^{\circ}\).

The line AO has the slope \(\displaystyle m_1=\frac{y_1}{x_1}\) and the line BO has the slope \(\displaystyle m_2=\frac{y_2}{x_2}\)

The tangent of the acute angle between the two lines is \(\displaystyle \tan\widehat{AOB}=\left|\displaystyle\frac{m_1-m_2}{1+m_1m_2}\right|\)

Replace \(\displaystyle m_1\) and \(\displaystyle m_2\):

\(\displaystyle 1=\tan 45^{\circ}=\left|\displaystyle\frac{x_2y_1-x_1y_2}{x_1x_2+y_1y_2}\right|\), (1)

The coordinates of A and B are the solutions of the system formed by the equation of the line and equation of the circle.

\(\displaystyle y=\displaystyle\frac{1-lx}{m}\).

Replace y in the equation of the circle and we have:

\(\displaystyle (m^2+l^2)x^2-2lx+1-a^2m^2=0\)

\(\displaystyle x_1, \ x_2\) are the roots of the quadratic and we have

\(\displaystyle x_1+x_2=\displaystyle\frac{2l}{m^2+l^2}, \ x_1x_2=\displaystyle\frac{1-a^2m^2}{m^2+l^2}\)

Replace \(\displaystyle y_1,y_2\) in (1):

\(\displaystyle 1=\displaystyle\frac{|m|\cdot|x_2-x_1|}{|(m^2+l^2)x_1x_2-l(x_1+x_2)+1|}=\frac{|m|\sqrt{(x_1+x_2)^2-4x_1x_2}}{|(m^2+l^2)x_1x_2-l(x_1+x_2)+1|}\).

Now replace \(\displaystyle x_1+x_2\) and \(\displaystyle x_1x_2\).

\(\displaystyle 1=\displaystyle\frac{2\sqrt{a^2m^2+a^2l^2-1}}{2-a^2m^2-a^2l^2}\Rightarrow 2\sqrt{a^2m^2+a^2l^2-1}=1-(a^2m^2+a^2l^2-1)\)

Let \(\displaystyle \sqrt{a^2m^2+a^2l^2-1}=t\).

Then \(\displaystyle 2t=1-t^2\Rightarrow t^2+2t-1=0\)

The positive root is \(\displaystyle t=\sqrt{2}-1\).

\(\displaystyle \sqrt{a^2m^2+a^2l^2-1}=\sqrt{2}-1\).

Square both members:

\(\displaystyle a^2m^2+a^2l^2-1=3-2\sqrt{2}\Rightarrow a^2(m^2+l^2)=4-2\sqrt{2}\)
 
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Opalg

MHF Hall of Honor
Aug 2007
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Leeds, UK
The circle has radius \(\displaystyle a\) and is centred at the origin. If a chord of the circle subtends an angle \(\displaystyle 45^\circ\) at the centre of the circle then the midpoint of the chord is at a distance \(\displaystyle d = a\cos22.5^\circ\) from the origin. But \(\displaystyle \cos^222.5^\circ = \frac{\sqrt2 +1}{2\sqrt2} = \frac1{4-2\sqrt2}\), as you can check from the formula \(\displaystyle \frac1{\sqrt2} = \cos45^\circ = 2\cos^222.5^\circ - 1\). It follows that \(\displaystyle d^2 = \frac{a^2}{4-2\sqrt2}\).

The line \(\displaystyle lx+my = 1\) intersects the circle at two points \(\displaystyle (x_1,y_1)\), \(\displaystyle (x_2,y_2)\) whose x-coordinates are the roots of the quadratic equation \(\displaystyle x^2 + (1 - lx)^2/m^2 = a^2\), or \(\displaystyle (l^2+m^2)x^2 - 2lx +1-a^2m^2 = 0\). The sum of the roots of this equation is \(\displaystyle x_1+x_2 = \frac {2l}{l^2+m^2}\). Thus the midpoint of the chord has x-coordinate \(\displaystyle x_{\text{mid}} = \frac12(x_1+x_2) = \frac {l}{l^2+m^2}\). Similarly, the midpoint of the chord has y-coordinate \(\displaystyle y_{\text{mid}} = \frac {m}{l^2+m^2}\). Thus the distance of the midpoint from the origin is given by \(\displaystyle d^2 = \frac{l^2+m^2}{(l^2+m^2)^2} = \frac1{l^2+m^2}.\)

Compare the two expressions for \(\displaystyle d^2\) to see that \(\displaystyle a^2(l^2+m^2) = 4-2\sqrt2\).
 
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