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Math Help - coordinate geometry regarding circle

  1. #1
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    coordinate geometry regarding circle

    Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.


    what i tried to do is:
    y=-2x+6 -----(1)
    let centre of circle be (x,y)
    (-2+x)^2+(0+y)^2=r^2
    4-4x+x^2+y^2=r^2 ------(2)
    (4+x)^2+(6+y)^2=r^2
    16+8x+x^2+36+12y+y^2=r^2 -----(3)
    Subst. (1) and (2) into (3),
    x=10
    Subst. x=10 into (1),
    y=-14
    centre of circle is (10,-14)..


    but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by wintersoltice View Post
    Find the equation of the circle which passes through the points P(-2,0), Q(4,6) and has its centre on the line 2x+y=6.


    what i tried to do is:
    y=-2x+6 -----(1)
    let centre of circle be (x,y)
    (-2+x)^2+(0+y)^2=r^2
    4-4x+x^2+y^2=r^2 ------(2)
    (4+x)^2+(6+y)^2=r^2
    16+8x+x^2+36+12y+y^2=r^2 -----(3)
    Subst. (1) and (2) into (3),
    x=10
    Subst. x=10 into (1),
    y=-14
    centre of circle is (10,-14)..


    but the answer is (2,2).....where did i go wrong? is my approach to the question even correct in the first place??
    I would do it like this:

    Let (x_0,y_0) be a point on the line (in this case, also let it be the center of our circle).

    Thus, the equation would be (x-x_0)^2+(y-y_0)^2=r^2

    WLOG, let r be the distance from (x_0,y_0) to (-2,0): r=\sqrt{(x_0+2)^2+y_0^2}

    So now it follows that (x-x_0)^2+(y-y_0)^2=(x_0+2)^2+y_0^2.

    Now, if we plug in the other point, we have

    (4-x_0)^2+(6-y_0)^2=(x_0+2)^2+(y_0)^2\implies 16-8x_0+x_0^2+36-12y_0+y_0^2=x_0^2+4x_0+4+y_0^2.

    Simplifying, we have 48=12x_0+12y_0\implies x_0+y_0=4

    Recall that (x_0,y_0) is on the line 2x+y=6\implies y_0=6-2x_0.

    Thus, x_0+(6-2x_0)=4\implies -x_0+6=4\implies -x_0=-2\implies x_0=2

    Therefore, y_0=6-2x_0=6-2(2)=6-4=2.

    Therefore, the center of the circle is (2,2) and the equation now becomes (x-2)^2+(y-2)^2=20

    Does this make sense?
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