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Math Help - Finding the equation of a circle

  1. #1
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    Finding the equation of a circle

    Find the equation of the circle, centre (2, -3), which touches the x axis.


    Please show all working out.
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  2. #2
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    Quote Originally Posted by Joker37 View Post
    Find the equation of the circle, centre (2, -3), which touches the x axis.


    Please show all working out.
    If the circle touches the x-axis, then the circle must pass through the point (2, 0). The distance from this point to the center (2, -3) is 3, so the radius is 3.

    Thus,
    \begin{aligned}<br />
(x - h)^2 + (y - k)^2 &= r^2 \\<br />
(x - 2)^2 + (y - (-3))^2 &= 3^2 \\<br />
(x - 2)^2 + (y + 3)^2 &= 9 \\<br />
\end{aligned}


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  3. #3
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    Quote Originally Posted by Joker37 View Post
    Find the equation of the circle, centre (2, -3), which touches the x axis.
    How far is the point (2,-3) from the x-axis?
    That will be the radius.
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  4. #4
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    Quote Originally Posted by yeongil View Post
    \begin{aligned}<br />
(x - 2)^2 + (y - (-3))^2 &= 3^2 \\<br />
\end{aligned}
    How did you get (y - (-3))^2? How come it's not (y - 3)^2
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  5. #5
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    Quote Originally Posted by Joker37 View Post
    How did you get (y - (-3))^2? How come it's not (y - 3)^2
    Because the y-coordinate is -3. In the equation of the circle:
    (x {\color{red}-} h)^2 + (y {\color{red}-} k)^2 = r^2
    It's y minus the y-coordinate of the center, or (y - (-3)), or (y + 3).


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