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Math Help - Conics - Distance between Two Points on Circle

  1. #1
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    Conics - Distance between Two Points on Circle

    R (x, y) P (a,b) Q (c,d) are points on x^2 + y^2 + 2gx + 2fy +k = 0

    i) if d is the distance between points R and P, show that:
    -d^2/2 = x.a + y.b + g(x+a) + f(y+b) + k

    I am not sure how to begin this question. Any approach is possible. Thanks guys
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    R (x, y) P (a,b) Q (c,d) are points on x^2 + y^2 + 2gx + 2fy +k = 0

    i) if d is the distance between points R and P, show that:
    -d^2/2 = x.a + y.b + g(x+a) + f(y+b) + k

    I am not sure how to begin this question. Any approach is possible. Thanks guys
    Mid point of RP is ( \frac{x+a}{2}, \frac{y+b}{2})

    If ( -g, -f ) is the center of the circle,

    then (\frac{d}{2})^2 = [(a+g)^2 + (b+f)^2] - [(\frac{x+a}{2} + g)^2 + (\frac{y+b}{2} + f)^2]

    Simplify and proceed.
    Last edited by sa-ri-ga-ma; June 5th 2010 at 06:49 AM.
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  3. #3
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    Could you just expand on how the centre of circle is acquired?

    Also, can i just acquire that the method used was by pythagoras? i.e. mid pt to centre is perpendicular to PR?

    Thanks.
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  4. #4
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    Quote Originally Posted by Lukybear View Post
    Could you just expand on how the centre of circle is acquired?

    Also, can i just acquire that the method used was by pythagoras? i.e. mid pt to centre is perpendicular to PR?

    Thanks.
    The general equation of the circle is x^2 + y^2 + 2gx + 2fy +K = 0

    The center of this circle is (-g, -f). If r is the radius of this circle, then

    (x+g)^2 + (y+f)^2 = r^2

    x^2 + 2gx + g^2 + y^2 + 2fy + f^2 = r^2

    x^2 + y^2 + 2gx + 2fy + g^2 + f^2 - r^2 = 0

    x^2 + y^2 + 2gx + 2fy + k = 0

    For the second part of the question, it is yes.
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  5. #5
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    Hello, Lukybear!

    Your variables are confusing.
    . . d is a coordinate . . . and a distance ?
    . . x is a coordinate . . . and a variable ?

    I must revise the problem.


    P(a,b)\text{ and }R(c,d) are points on: . x^2 + y^2 + 2px + 2qy +r \:=\: 0

    i) If D is the distance between points P and R, show that:

    . . -\frac{D^2}{2} \;=\;ac + bd + p(a+c) + q(b+d) + r


    \begin{array}{ccccccc}P(a,b)\text{ is on the circle:} &a^2+b^2+2pa + 2qb + r &=& 0 \\ \\[-3mm]<br />
R(c,d)\text{ is on the circle:} & c^2+d^2+2pc + 2qd + r &=& 0 \end{array}


    Add the equations: . a^2+c^2+b^2+d^2+2pa + 2pc + 2qb + 2qd + 2r \:=\:0

    And we have: . (a^2+c^2)+(b^2+d^2) + 2p(a+c) + 2q(b+d) + 2r \:=\:0

    . . Hence: . (a^2+b^2) + (b^2+d^2) \;=\;-2\bigg[p(a+c) + q(b+d) + r\bigg] .[1]


    The distance between P and R is given by: . D^2 \;=\;(a-c)^2 + (b-d)^2)

    And we have: . D^2 \;=\;a^2-2ac + c^2 + b^2 - 2bd + d^2

    . . Hence: . D^2 \;=\;(a^2+c^2) + (b^2 + d^2) - 2(ac + bd) .[2]


    Substitute [1] into [2]: . D^2 \;=\;-2\bigg[p(a+c) + q(b+d) + r\bigg] - 2(ac+bd)

    And we have: . D^2 \;=\;-2(ac+bd) - 2p(a+c) - 2q(b+d) - 2r

    . . Therefore: . -\frac{D^2}{2} \;=\;ac+bd + p(a+c) + q(b+d) + r

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  6. #6
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    Wow thats brilliant. Thxs very much! Do apologise for question.
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