math

Find the centre of the circle that passes through the points p(-9,5), q(1,5), and r(-2,-2)

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asked by sally
  1. consider the two chords PQ and PR
    the centre must be the intersection of the right-bisector of these two chords.

    midpoint of PQ= (-4,5)
    slope of PQ = 0
    equation of right-bisector of PQ is x = -4
    (no real work required for this one, notice that PQ was a horizontal line, so the right-bisector must be the vertical line running through its midpoint)

    midpoint of PR = (-11/2 , 3/2)
    slope of PR = -7/7 = -1
    so slope of perpendicular = +1
    equation: x - y = C
    but (-11/2 , 3/2) is on it
    -11/2 -3/2 =c = -7
    x - y = -7

    sub in x = -4
    -4 - y = -7
    -y = -3
    y = 3

    the centre is (-4,3)

    method2:

    let the centre be (a,b)

    the distance from (a,b) to each of those points must be the same ..., so

    √((a+9)^2 + (b-5)^2) = √((a-1)^2 + (b-5)^2)
    a^2 + 18a + 81 + b^2 - 10b + 25 = a^2 - 2a + 1 + b^2 -10b + 25
    20a = -80
    a = -4 , as before

    √( (a+9)^2 + (b-5)^2) = √(a+2)^2 + (b+2)^2)
    a^2 + 18a + 81 + b^2 - 10b + 25 = a^2 + 4a + 4 + b^2 + 4b + 4
    14a -14b = -98
    a - b = -7
    but a=-4
    -4 - b = -7
    b = 3

    then centre is (-4,3) , as before

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    posted by Reiny

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