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Thread: co-ordinate geometry

  1. #1
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    Question co-ordinate geometry

    points (1,2),(3,-4),(5,-6) laying on the circumference of a circle,find the cordinate of centre?
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  2. #2
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    Quote Originally Posted by anjana View Post
    points (1,2),(3,-4),(5,-6) laying on the circumference of a circle,find the cordinate of centre?
    Hello, Anjana,

    I assume that the equation of a circle in $\displaystyle \mathbb{R}^2$ with the centre $\displaystyle M(m_1, m_2)$ is:
    $\displaystyle (x-m_1)^2+(y-m_2)^2=r^2$

    Now plug in the values yxou know. You'll get 3 equations:

    $\displaystyle (1-m_1)^2+(2-m_2)^2=r^2$
    $\displaystyle (3-m_1)^2+(-4-m_2)^2=r^2$
    $\displaystyle (5-m_1)^2+(-6-m_2)^2=r^2$

    Expand the LHS of these equations. Then substract equ2 from equ1 and equ3 from equ1. You'll get 2 linear equations without any squares which will give you the values of m1 and m2. Afterwards you can calculate the value of r.

    For confirmation only M(11, 2) and r = 10

    EB
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  3. #3
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    Hello, anjana!

    EB has the best solution.
    Here's another approach . . . requiring more Thinking.


    Points $\displaystyle A(1,2),\:B(3,-4),\:C(5,-6)$ are on the circumference of a circle.
    Find the coordinates of the centre.

    The circumcenter is the intersection of the perpendicular bisectors of the sides.

    The midpoint of $\displaystyle AB$ is $\displaystyle (2,-1)$. The slope of $\displaystyle AB$ is: $\displaystyle \frac{-4-2}{3-1} = -3$
    The equation of the perpendicular bisector of $\displaystyle AB$ is:
    . . $\displaystyle y - (-1) \:=\:\frac{1}{3}(x - 2)\quad\Rightarrow\quad y\:=\:\frac{1}{3}x - \frac{5}{3}$

    The midpoint of $\displaystyle BC$ is $\displaystyle (4,-5).$ The slope of $\displaystyle BC$ is $\displaystyle \frac{-6+4}{5-3}\:=\:-1$
    The equation of the perpendicular bisector of $\displaystyle BC$ is:
    . . $\displaystyle y- (-5) \:=\:1(x-4)\quad\Rightarrow\quad y \:=\:x - 9$


    The perpendicular bisectors intersect when:
    . . $\displaystyle \frac{1}{3}x - \frac{5}{3}\:=\:x - 9\quad\Rightarrow\quad\boxed{x\,=\,11}$

    Then: .$\displaystyle y \:=\:11 - 9\quad\Rightarrow\quad\boxed{y \,=\,2}$


    Therefore, the center of the circle is: .$\displaystyle (11,\,2)$

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