# Thread: Finding the equation of a circle

1. ## Finding the equation of a circle

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

2. Originally Posted by Joker37
Find the equation of the circle, centre (2, -3), which touches the x axis.

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}
(x - h)^2 + (y - k)^2 &= r^2 \\
(x - 2)^2 + (y - (-3))^2 &= 3^2 \\
(x - 2)^2 + (y + 3)^2 &= 9 \\
\end{aligned}

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3. Originally Posted by Joker37
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?

4. Originally Posted by yeongil
\begin{aligned}
(x - 2)^2 + (y - (-3))^2 &= 3^2 \\
\end{aligned}
How did you get (y - (-3))^2? How come it's not (y - 3)^2

5. Originally Posted by Joker37
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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