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Math Help - 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 A(x_1,y_1), \ B(x_1,y_1) be the intersect points between the line and the circle. Then \widehat{AOB}=45^{\circ}.

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

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

    Replace m_1 and m_2:

    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.

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

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

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

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

    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 y_1,y_2 in (1):

    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 x_1+x_2 and x_1x_2.

    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 \sqrt{a^2m^2+a^2l^2-1}=t.

    Then 2t=1-t^2\Rightarrow t^2+2t-1=0

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

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

    Square both members:

    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; July 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 a and is centred at the origin. If a chord of the circle subtends an angle 45^\circ at the centre of the circle then the midpoint of the chord is at a distance d = a\cos22.5^\circ from the origin. But \cos^222.5^\circ = \frac{\sqrt2 +1}{2\sqrt2} = \frac1{4-2\sqrt2}, as you can check from the formula \frac1{\sqrt2} = \cos45^\circ = 2\cos^222.5^\circ - 1. It follows that d^2 = \frac{a^2}{4-2\sqrt2}.

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

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