Finding the equation of a circle

**Joker37** · June 20th 2009, 04:39 AM

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

Please show all working out.

**yeongil** · June 20th 2009, 04:55 AM

Originally Posted by Joker37

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

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**Plato** · June 20th 2009, 04:56 AM

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?
That will be the radius.

**Joker37** · June 20th 2009, 05:06 AM

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

**yeongil** · June 20th 2009, 05:27 AM

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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