# Thread: Equation of a Circle

1. ## Equation of a Circle

Find the equation of a circle whose centre falls on the line y = 6 - 2x and which passes through the points A(-2,0) and B(4,0)
Since the y-value of both points A and B are 0, I can gather from this question that the diameter is 6 units and so the radius is 3 units. But I can't solve any further.
Any help would be greatly appreciated, thank you!
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2. Well the equation of a circle with radius size $\displaystyle r$ is $\displaystyle \displaystyle (x-h)^2+(y-k)^2=r^2$ where $\displaystyle (h,k)$ is its centre.

Now use the information given (substitute in some points) to solve for $\displaystyle r, h, k$

3. Hello, caramelcake!

Find the equation of the circle whose center is on $\displaystyle y \:=\:6-2x$
and which passes through $\displaystyle A(\text{-}2,0)$ and $\displaystyle B(4,0).$

Since the $\displaystyle y$-value of both points $\displaystyle A$ and $\displaystyle B$ are 0,
I can gather that the diameter is 6 units. . This is not true!

Did you make a sketch?

Code:
           \    |
\   * * *
*\  |       *
*   \ | :       *
*     \| :        *
* :
*       |\:C        *
*       | o         *
*       | :\        *
| : \
*      | :  \     *
- - o - - + - - \ - o - - - - -
-2 *   | :    \* 4
* * *   \
| :

The $\displaystyle x$-intercepts of the circle are: $\displaystyle A(\text{-}2,0)$ and $\displaystyle B(4,0)$

The center lies on the vertical line midway between the intercepts: .$\displaystyle x \:=\:1$

The center also lies on the line $\displaystyle y \:=\:6-2x$

. . The center is the intersection of the two lines: .$\displaystyle C(1,4)$

The radius is the distance $\displaystyle \overline{CA}\!:\;5$

Therefore: .$\displaystyle (x-1)^2 + (y-4)^2 \:=\:5^2$

4. Thank you pickslides and Soroban for your input Terribly sorry for the late reply, I didn't log in in time to check the second reply before submitting my homework >.<
x-coordinate of centre is $\displaystyle \frac{-2+4}{2}$$\displaystyle =1$
$\displaystyle y = 6 - 2 = 4$
radius of circle = $\displaystyle \sqrt{(1 + 2)^2 + (4 - 0)^2}$
$\displaystyle = \sqrt{25}$
$\displaystyle = 5 units$
$\displaystyle (x + 1)^2 + (y - 4)^2 = 5^2$
$\displaystyle x^2 - 2x +1 + y^2 - 8y + 16 = 25$
$\displaystyle x^2 + y^2 - 2x - 8y -8 = 0$