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Thread: Finding Circle

  1. #1
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    Finding Circle

    I'm having hard time finding equation of a circle with given two points and a radius.

    given points (-2,3) and (2,-1) and radius = 4

    my midpoint is
    $\displaystyle (\frac{-2+2}{2},\frac{3-1}{2})$
    =(0,1)

    my distance is :

    $\displaystyle d=\sqrt{(2+2)^2+(-1-3)^2}$
    $\displaystyle =\sqrt{32}$

    What should I do next to find the center and the equation of the circle?

    thanks in advance:
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Anemori View Post
    I'm having hard time finding equation of a circle with given two points and a radius.

    given points (-2,3) and (2,-1) and radius = 4

    my midpoint is
    $\displaystyle (\frac{-2+2}{2},\frac{3-1}{2})$
    =(0,1)

    my distance is :

    $\displaystyle d=\sqrt{(2+2)^2+(-1-3)^2}$
    $\displaystyle =\sqrt{32}$

    What should I do next to find the center and the equation of the circle?

    thanks in advance:
    Let's assume that $\displaystyle k:\; (x-u)^2+(y-v)^2=16$ is the equation of the circle with center $\displaystyle C(u,v)$ and radius 4.

    Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

    $\displaystyle (-2-u)^2+(3-v)^2=16$ and $\displaystyle (2-u)^2+(-1-v)^2=16$.
    If you now subtract the second equation from the first, you get $\displaystyle 8u-8v+8=0$, and hence $\displaystyle u=v-1$.

    Replacing $\displaystyle u$ in the first equation by $\displaystyle v-1$ gives you the quadratic equation $\displaystyle (-1-v)^2+(3-v)^2=16$.
    Now solve for $\displaystyle v$: this gives you two solutions $\displaystyle v_1,v_2$ for $\displaystyle v$. Finally, pair these two solutions with the corresponding x-coordinates of C, namely $\displaystyle u_{1,2}=v_{1,2}-1$, and you are done.
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  3. #3
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    Quote Originally Posted by Failure View Post
    Let's assume that $\displaystyle k:\; (x-u)^2+(y-v)^2=16$ is the equation of the circle with center $\displaystyle C(u,v)$ and radius 4.

    Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

    $\displaystyle (-2-u)^2+(3-v)^2=16$ and $\displaystyle (2-u)^2+(-1-v)^2=16$.
    If you now subtract the second equation from the first, you get $\displaystyle 8u-8v+8=0$, and hence $\displaystyle u=v-1$.

    Replacing $\displaystyle u$ in the first equation by $\displaystyle v-1$ gives you the quadratic equation $\displaystyle (-1-v)^2+(3-v)^2=16$.
    Now solve for $\displaystyle v$: this gives you two solutions $\displaystyle v_1,v_2$ for $\displaystyle v$. Finally, pair these two solutions with the corresponding x-coordinates of C, namely $\displaystyle u_{1,2}=v_{1,2}-1$, and you are done.

    im lost where u=v-1...
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Anemori View Post
    im lost where u=v-1...
    Well, as I wrote, you can now replace u in $\displaystyle (-2-u)^2+(3-v)^2=16$ by v-1.

    This gives you that $\displaystyle (-1-v)^2+(3-v)^2=16$ must hold. Thus it follows from this quadratic equation that $\displaystyle v_1=-1$ or $\displaystyle v_2=3$.

    Since we know that $\displaystyle u=v-1$, we get for the corresponding x-coordinates $\displaystyle u_{1,2}$ of the center of the circle, that $\displaystyle u_1=v_1-1=-2$ and $\displaystyle u_2=v_2-1=2$, respectively.

    Thus the center of the circle is either $\displaystyle C_1(-2,-1)$ or $\displaystyle C_2(2,3)$, and the corresponding equations of these two circles are

    $\displaystyle k_1:\; (x+2)^2+(y+1)^2=4^2$ and $\displaystyle k_2:\; (x-2)^2+(y-3)^2=4^2$.
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