1. ## coordinate circle

A circle in the standard (x,y) coordinate plane has center (3,8) and is tangent to the x-axis. The point (x,y) in on the circle if and only if x and y satisfy which of the follwing equations?

The solution is (x - 3)^2 + (y - 8)^2 = 64. Why do we say the radius is 8? Why couldn't it be 3 with the above information?

2. Originally Posted by sarahh
A circle in the standard (x,y) coordinate plane has center (3,8) and is tangent to the x-axis. The point (x,y) in on the circle if and only if x and y satisfy which of the follwing equations?

The solution is (x - 3)^2 + (y - 8)^2 = 64. Why do we say the radius is 8? Why couldn't it be 3 with the above information?
As you may know the radius is perpendicular to the tangent in the tangent point. The tangent point is T(3, 0) because it is placed on the x-axis and therefore $y_T = 0$ and it must be perpendicular below the center and therefore $x_T = 3$.

Now calculate the distance between these 2 points:

$r = \sqrt{(3-3)^2+(8-0)^2} = 8$

3. Hello, Sarah!

Did you make a sketch?

A circle in the standard (x,y) coordinate plane
has center (3,8) and is tangent to the x-axis.
The point (x,y) in on the circle if and only if x and y satisfy which of the follwing equations?

The solution is (x - 3)^2 + (y - 8)^2 = 64.
Why do we say the radius is 8?
Why couldn't it be 3 with the above information?
Code:
          |
|   * * *
*           *
* |             *
*  |              *
|
*   |   (3,8)       *
*   |     *         *
*   |     :         *
|     :
*  |     :        *
* |     :       *
*     :     *
- - - + - * * * - - - - - -
|