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Thread: 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 $\displaystyle \sqrt{m^2+1}$, & why it's simply multiplied against $\displaystyle |3m+4|$? After multiplying that expression, I end up with $\displaystyle (3m^2+24m+16)(m^2+1)=20$, & not $\displaystyle (11m-2)(m-2)=0$. Can anyone help clarify this for me, or is there a better approach?

    Many thanks.

    Q. $\displaystyle x^2+y^2+2x-4y-20=0$ 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 $\displaystyle 2\sqrt{5}$. Find the equation of L & M.

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

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

    Ans. (From text book): $\displaystyle 2x-y-6=0$ & $\displaystyle 2x-11y-26=0$
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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 $\displaystyle \sqrt{m^2+1}$, & why it's simply multiplied against $\displaystyle |3m+4|$? After multiplying that expression, I end up with $\displaystyle (3m^2+24m+16)(m^2+1)=20$, & not $\displaystyle (11m-2)(m-2)=0$. Can anyone help clarify this for me, or is there a better approach?
    Attempt: Line equation for (2,-2): $\displaystyle y+2=m(x-2)$
    From circle equation: centre c = (-1, 2)
    Sub c into the line equation: $\displaystyle 2+2=m(-1-2)\rightarrow 3m+4=0$

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

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


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

    So the distance is $\displaystyle \frac{|-m-2-2-2m|}{\sqrt{m^2+1}}=2\sqrt{5}$ or $\displaystyle \frac{(3m+4)^2}{m^2+1}}=20$
    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 $\displaystyle (-1,2)$ and has a radius of 5. If we draw a second smaller circle with the same centre and with radius $\displaystyle 2\sqrt{5},$ then what we are looking for are the two tangents from the point $\displaystyle (2,-2),$ (which incidently lies on the larger circle), to the smaller one.

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

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

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

    The equations of the two tangents can then be calculated (and agree with the given answers).
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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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