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Math Help - Circles and Chords

  1. #1
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    Circles and Chords

    Below, I have posted a question and a solution I was provided. The problem is that I don't understand how the part bordered by AAA was arrived at. Specifically \sqrt{m^2+1}, & why it's simply multiplied against |3m+4|? After multiplying that expression, I end up with (3m^2+24m+16)(m^2+1)=20, & not (11m-2)(m-2)=0. Can anyone help clarify this for me, or is there a better approach?

    Many thanks.

    Q. x^2+y^2+2x-4y-20=0 is the equation of a circle. 2 lines, L & M, intersect at (2, -2). The distance from the centre of the circle to each line is 2\sqrt{5}. Find the equation of L & M.

    Attempt: Line equation for (2,-2): y+2=m(x-2)
    From circle equation: centre c = (-1, 2)
    Sub c into the line equation: 2+2=m(-1-2)\rightarrow 3m+4=0

    AAA
    The distance from the point (-1, 2) to this line is |3m+4|\sqrt{m^2+1}=2\sqrt{5}
    (11m-2)(m-2)=0
    AAA

    Ans. (From text book): 2x-y-6=0 & 2x-11y-26=0
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  2. #2
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    Re: Circles and Chords

    Quote Originally Posted by GrigOrig99 View Post
    Below, I have posted a question and a solution I was provided. The problem is that I don't understand how the part bordered by AAA was arrived at. Specifically \sqrt{m^2+1}, & why it's simply multiplied against |3m+4|? After multiplying that expression, I end up with (3m^2+24m+16)(m^2+1)=20, & not (11m-2)(m-2)=0. Can anyone help clarify this for me, or is there a better approach?
    Attempt: Line equation for (2,-2): y+2=m(x-2)
    From circle equation: centre c = (-1, 2)
    Sub c into the line equation: 2+2=m(-1-2)\rightarrow 3m+4=0

    AAA
    The distance from the point (-1, 2) to this line is |3m+4|\sqrt{m^2+1}=2\sqrt{5}
    (11m-2)(m-2)=0AAA

    Ans. (From text book): 2x-y-6=0 & 2x-11y-26=0


    The equations of the lines are mx-y-2-2m=0.

    So the distance is \frac{|-m-2-2-2m|}{\sqrt{m^2+1}}=2\sqrt{5} or \frac{(3m+4)^2}{m^2+1}}=20
    Thanks from GrigOrig99
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  3. #3
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    Re: Circles and Chords

    Thank you very much.
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  4. #4
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    Re: Circles and Chords

    I don't think that this is necessarily a better approach, but it is different.

    The circle has a centre at (-1,2) and has a radius of 5. If we draw a second smaller circle with the same centre and with radius 2\sqrt{5}, then what we are looking for are the two tangents from the point (2,-2), (which incidently lies on the larger circle), to the smaller one.

    The smaller circle will have equation (x+1)^{2}+(y-2)^{2}=(2\sqrt{5})^{2}, or,
    x^{2}+y^{2}+2x-4y-15=0,
    and a tangent to this circle at the point (x_{1},y_{1}) will have equation
    xx_{1}+yy_{1}+(x+x_{1})-2(y+y_{1})-15=0.

    This, (these), has (have) to pass through the point (2,-2), so 3x_{1}-4y_{1}-9=0.

    The point (x_{1},y_{1}) lies on the smaller circle, so substituting x_{1}=(4y_{1}+9)/3, allows us to find the y co-ordinates of the two points of contact. These turn out to be 0 \text{ and } -12/5.
    Corresponding x co-ordinates are 3 \text{ and } -1/5.

    The equations of the two tangents can then be calculated (and agree with the given answers).
    Thanks from GrigOrig99
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  5. #5
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    Re: Circles and Chords

    Circles and Chords-23-may-13.png
    Thanks from GrigOrig99
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