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Math Help - question to with geometry of a circle

  1. #1
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    question to with geometry of a circle

    Hello i need help with the following question

    I have attached my question

    thanks

    Kind Regards
    Attached Thumbnails Attached Thumbnails question to with geometry of a circle-question.jpg  
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  2. #2
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    The centre of the circle must lie on the perpendicular bisector of PQ. This is the line through the midpoint (3,1) of PQ, with gradient 2. So its equation is y = 2x-7.

    If a circle with centre at the point (x,2x-7) touches the y-axis then its radius must be equal to x. If that circle passes through the point (1,0) then (x-1)^2 + (2x-7)^2 (the square of the distance from (x,2x-7) to (1,0)) must therefore be equal to x^2.

    Solve the equation (x-1)^2 + (2x-7)^2 = x^2 to find the x-coordinate of the centre of the circle (two solutions, because it is a quadratic equation).
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  3. #3
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    Talking

    FYI for other viewers: The text of the exercise (from the graphic) is as follows:

    Two circles touch the y-axis and pass through the points Q(1, 0) and P(5, -2). Find the equations of the two circles and determine their radii and centers.
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  4. #4
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    Hello, cooltowns!

    Two circles touch the y-axis and pass through the points Q(1, 0) and P(5, -2).
    Find the equations of the two circles and determine their radii and centers.

    We want to locate the center of the circle C(x,y)
    . . which is equidistant from P, Q and the y-axis.
    Code:
            |
            |           C
          Y o  *  *  *  o (x,y)
            |         *  *
            |       *     *
            |   Q *        *
        - - + - o - - - - - * - - - -
            | (1,0)          o P
            |              (5,-2)

    We want: .  CP \,=\,CQ \,=\,CY

    \text{Then: }\;\underbrace{\sqrt{(x-5)^2 + (y-2)^2}}_{[1]} \;=\;\underbrace{\sqrt{(x-1)^2 + y^2}}_{[2]} \;=\;\underbrace{x}_{[3]}


    Square [2] and [3]: . (x-1)^2 + y^2 \:=\:x^2 \quad\Rightarrow\quad x\:=\:\frac{y^2+1}{2} .(a)

    Square [1] and [3]: . (x-5)^2 + (y-2)^2 \:=\:x^2 \quad\Rightarrow\quad y^2 + 4y + 29 \:=\:10x .(b)

    Substitute (a) into (b): . y^2 + 4y + 29 \:=\:10\left(\frac{y^2+1}{2}\right) \quad\Rightarrow\quad y^2-y-6\:=\:0

    Then: . (y-3)(y+2)\:=\:0 \quad\Rightarrow\quad y \:=\:3,\:\text{-}2 \quad\Rightarrow\quad x \:=\:5,\:\tfrac{5}{2}


    The circles are: . \begin{Bmatrix}(x-5)^2 + (y-3)^2 \:=\:25 & C(5,3) & r = 5 \\ \\[-3mm]<br />
\left(x-\frac{5}{2}\right)^2 + (y+2)^2 \:=\:\frac{25}{4} & C\left(\frac{5}{2},\:\text{-}2\right) & r = \frac{5}{2} \end{Bmatrix}


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