http://library.thinkquest.org/20991/...2_circles2.gif **Circles, when graphed on the coordinate plane, have an equation of ***x2 + y2 = r2* where *r* is the radius (standard form) when the center of the circle is the origin. When the center of the circle is *(h, k)* and the radius is of length *r*, the equation of a circle (standard form) is *(x - h)2 + (y - k)2 = r2*. **Example: ****1. Problem: Find the center and radius of ***(x - 2)sq + (y + 3)sq = 16*. Then graph the circle. **Solution:** Rewrite the equation in standard form.

*(x - 2)sq + [y - (-3)]sq = 4sq*
The center is

*(2, -3)* and the radius is

*4*. The graph is easy to draw, especially if you use a compass. The figure below is the graph of the solution.

http://library.thinkquest.org/20991/...g2_circles.gif
here are my questions:

~How did they get 2 and -3 as the center point of the circle?

**Answer: The equation of a circle is Pythagoras' theorem applied to every point (x,y) on the circle circumference.**

Pick any point on the circumference.

Using the centre, draw a right-angled triangle.

The radius of the circle is the hypotenuse of the right-angled triangle.

The length of the horizontal side is the difference between the x co-ordinates.

The length of the vertical side is the difference between the y co-ordinates.

Therefore Pythagoras' theorem gives
$\displaystyle \left(x-x_c\right)^2+\left(y-y_c\right)^2=r^2$

$\displaystyle \left(x_c,y_c\right)$

**is the circle centre.**
~How come the formula change from x2+y2= r2 to (x-h)2 + (y-k)2 = R2?

**Answer: If the centre is (0,0), then Pythagoras' theorem gives**
$\displaystyle (x-0)^2+(y-0)^2=x^2+y^2=r^2$

~Sorry i'm kind of dumb on these things. =) thank you