# Thread: Equation of circles help?

1. ## Equation of circles help?

The centre of two circles lie on the line x+y-7=0

Both circles touch the x - axis and contain the point (6,2), find the equations of the two circles that satisfy these conditions.

2. Hello, sam1234!

The centres of two circles lie on the line $\displaystyle x+y \:=\:7$
Both circles touch the x-axis and contain the point $\displaystyle P(6,2).$

Find the equations of the two circles.
Code:
            |
*       | * * *
*   * |         *
*   |           *
*  * |            *
*
*     | * C         *
*     |   o (h,k)   *
*     |   : *       *
|   :   *
*    |   :k    *  o P(6,2)
*   |   :       *
------*-+---+-----*---*------
| * * *         *
|

The center is $\displaystyle C(h,k)$, which lies on $\displaystyle x+y-7 \:=\:0$
. . Hence: .$\displaystyle h+k-7 \:=\:0 \quad\Rightarrow\quad k \:=\:7-h$ .[1]

The circle is tangent to the $\displaystyle x$-axis, so the radius is $\displaystyle k.$

The equation is: .$\displaystyle (x-h)^2 + (y-k)^2 \:=\:k^2$

Since $\displaystyle (6,2)$ is on the circle: .$\displaystyle (6-h)^2 + (2-k)^2 \:=\:k^2 \quad\Rightarrow\quad h^2-12h + 40 - 4k \:=\:0$ .[2]

Substitute [1] into [2]: .$\displaystyle h^2-12h + 40 - 4(7-h) \:=\:0 \quad\Rightarrow\quad h^2-8h + 12 \:=\:0$

Then: .$\displaystyle (h-2)(h-6) \:=\:0 \quad\Rightarrow\quad h \;=\;2,\:6$

. . Hence: .$\displaystyle k \;=\;5,\:1$

Therefore: . $\displaystyle \begin{Bmatrix} C(2,5),\:r=5 && \Rightarrow &&(x-2)^2 + (y-5)^2 &=& 25 \\ \\[-3mm] C(6,1),\:r=1 && \Rightarrow && (x-6)^2 + (y-1)^2 &=& 1 \end{Bmatrix}$