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Thread: coordinate geometry

  1. #1
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    coordinate geometry

    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
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  2. #2
    MHF Contributor red_dog's Avatar
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    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:

    $\displaystyle a^2m^2+a^2l^2-1=3-2\sqrt{2}\Rightarrow a^2(m^2+l^2)=4-2\sqrt{2}$
    Last edited by mr fantastic; Jul 15th 2010 at 04:10 PM. Reason: Fixed a latex tag.
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  3. #3
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    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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