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Math Help - equation of circle question leading to parrallelograms

  1. #1
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    equation of circle question leading to parrallelograms

    Hi I need help with the following question

    I have snapshotted the question at attached it to this thread.



    any help is appreichated

    thank you
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    Hello, rpatel!

    Here's the first part.
    I'll try to get back to the rest later . . .


    (a) Find the equation of the circle that passes through A(2,3), B(0,-1), C(-4,-3).

    What is the center and radius of this circle?
    There are a number of approaches to this problem.
    I think this one is the simplest. (I may be the only one!)


    The center is the point P(x,y) equidistant from A,B,C.


    Equate the distances: . PA \:=\:PB \:=\:PC

    . . \underbrace{\sqrt{(x-2)^2+(y-3)^2}}_{{\color{blue}[1]}} \:=\:\underbrace{\sqrt{x^2+(y+1)^2}}_{{\color{blue  }[2]}} \:=\:\underbrace{\sqrt{(x+4)^2 + (y+3)^2}}_{{\color{blue}[3]}}


    From {\color{blue}[1] = [2]}, we have: . x^2-4x+4 + y^2-6y+9 \:=\:x^2+y^2+2y+1 \quad\Rightarrow\quad x + 2y \:=\:3

    From {\color{blue}[2] = [3]}, we have: . x^2+y^2+2y+1 \:=\:x^2+8x+16 + y^2+6y+9 \quad\Rightarrow\quad 2x + y \:=\:\text{-}6

    Solve the system of equations: . x \:=\:\text{-}5,\;y \:=\:4

    . . Hence, the center is: . P(-5,4)


    The radius is the distance PA ... (or  PB or PC).

    . . r \;=\;PA \:=\:\sqrt{(\text{-}5-2)^2 + (4-3)^2} \:=\: \sqrt{49+1} \:=\:\sqrt{50} \:=\:5\sqrt{2}


    Therefore, the equation of the circle is:

    . . \left(x-[\text{-}5]\right)^2 + (y - 4)^2 \:=\:(5\sqrt{2})^2 \quad\Rightarrow\quad\boxed{ (x+5)^2 + (y-4)^2 \:=\:50}

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  3. #3
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    Quote Originally Posted by rpatel View Post
    Hi I need help with the following question

    I have snapshotted the question at attached it to this thread.



    any help is appreichated

    thank you
    to c)

    There are 6 parallelograms:

    Take 2 points to determine one side of the parallelogram. Then you'll get the 4th vertex if you add to the staionary vector pointing to the 3rd vertex the vector which is defined by the 1st and 2nd vertex:

    The side is CB. Then you get the coordinates of D:

    \overrightarrow{OA}+\overrightarrow{CB}=\overright  arrow{OD_1}

    or

    \overrightarrow{OA}-\overrightarrow{CB}=\overrightarrow{OD_2}

    You can repeat this procedure with two more sides.
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  4. #4
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    Hello again, rpatel!

    (b) Find the equation of the circle with center (10,10)
    that just touches the circle you found in part (a).
    I won't try to draw a diagram for this. .I hope you can visualize my reasoning.


    The first circle is: . (x+5)^2 + (y-4)^2 \:=\:50
    Its center is: P(\text{-}5,4) and its radius is: 5\sqrt{2}.

    The second circle has center Q(10,10) and we must find its radius.


    The distance PQ is: . \sqrt{(10-[\text{-}5])^2 + (10-4)^2} \:=\:\sqrt{225+36} \:=\:\sqrt{261} \:=\:3\sqrt{29}

    . . Hence, the radius of the second circle is: . 3\sqrt{29}-5\sqrt{2}


    The equation is: . (x-10)^2 + (y-10^2 \:=\:(3\sqrt{29} -5\sqrt{2})^2

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