# coordinate geometry

#### thorpelizts

##### New member
find the equation of a circle whose center falls ont he line y=6-2x and which passes through the points A(-2,0) and B(4,0).

poor in circles. how to even start?

#### Sudharaka

##### Well-known member
MHB Math Helper
find the equation of a circle whose center falls ont he line y=6-2x and which passes through the points A(-2,0) and B(4,0).

poor in circles. how to even start?
Hi thorpelizts,

So the center of the circle should be $$(x_{0},6-2x_{0})$$. The equation of a circle with radius $$r$$ and center $$(a,b)$$ can be represented in Cartesian coordinates by,

$(x-a)^2+(y-b)^2=r^2$

In our case,

$(x-x_{0})^2+(y-6+2x_{0})^2=r^2$

Now we know that, $$A\equiv (-2,0)$$ and $$B\equiv (4,0)$$ lies on the circle. So these two points must satisfy the above equation. Then you will have two equations with two unknowns$$(r\mbox{ and }x_{0})$$. Hope you can continue.

Kind Regards,
Sudharaka.

#### soroban

##### Well-known member
Hello, thorpelizts!

Another approach . . .

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

Code:
             \|
6*
|\
| \
|  \
(0,4)oB  \
/|    \
/ |     \
*  |      \
/   *       \
A /    |  *     \3
- - o - - + - - * - + - - -
(-2,0)  |        * \
|           oC
|            \ *
|
The center lies on the line $$y \,=\,6-2x$$
The center lies on the perpendicular bisector of $$AB.$$
. . The center is the intersection of these two lines.

The midpoint of $$AB$$ is $$(\text{-}1,2)$$
The slope of $$AB$$ is 2.
The perpendicular slope is: $$\text{-}\tfrac{1}{2}$$
The equation of the perpendicular bisector is:
. . $$y - 2 \;=\;\text{-}\tfrac{1}{2}(x + 1) \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{1}{2}x + \tfrac{3}{2}$$

It has an x-intercept at (3,0).
And so does the other line!

Their intersection (and hence the center) is: $$C(3,0).$$

The radius is: $$AC = BC = 5.$$

Got it?

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi thorpelizts,

So the center of the circle should be $$(x_{0},6-2x_{0})$$. The equation of a circle with radius $$r$$ and center $$(a,b)$$ can be represented in Cartesian coordinates by,

$(x-a)^2+(y-b)^2=r^2$

In our case,

$(x-x_{0})^2+(y-6+2x_{0})^2=r^2$

Now we know that, $$A\equiv (-2,0)$$ and $$B\equiv (4,0)$$ lies on the circle. So these two points must satisfy the above equation. Then you will have two equations with two unknowns$$(r\mbox{ and }x_{0})$$. Hope you can continue.

Kind Regards,
Sudharaka.
Since a complete answer had been posted to the question let me complete my method,

$(-2-x_{0})^2+(-6+2x_{0})^2=r^2=(4-x_{0})^2+(-6+2x_{0})^2$

$\Rightarrow (-2-x_{0})^2=(4-x_{0})^2$

$\Rightarrow 4+4x_{0}=16-8x_{0}$

$\Rightarrow 12x_{0}=12$

$\Rightarrow x_{0}=1$

Therefore,

$r^2=(4-1)^2+(-6+2)^2=9+16=25$

$\Rightarrow r=5$

Kind Regards,
Sudharaka.

#### Sudharaka

##### Well-known member
MHB Math Helper
Hello, thorpelizts!

Another approach . . .

Code:
             \|
6*
|\
| \
|  \
(0,4)oB  \
/|    \
/ |     \
*  |      \
/   *       \
A /    |  *     \3
- - o - - + - - * - + - - -
(-2,0)  |        * \
|           oC
|            \ *
|
The center lies on the line $$y \,=\,6-2x$$
The center lies on the perpendicular bisector of $$AB.$$
. . The center is the intersection of these two lines.

The midpoint of $$AB$$ is $$(\text{-}1,2)$$
The slope of $$AB$$ is 2.
The perpendicular slope is: $$\text{-}\tfrac{1}{2}$$
The equation of the perpendicular bisector is:
. . $$y - 2 \;=\;\text{-}\tfrac{1}{2}(x + 1) \quad\Rightarrow\quad y \:=\:\text{-}\tfrac{1}{2}x + \tfrac{3}{2}$$

It has an x-intercept at (3,0).
And so does the other line!

Their intersection (and hence the center) is: $$C(3,0).$$

The radius is: $$AC = BC = 5.$$

Got it?
Hi soroban,

I think there is a slight mistake here. You have taken $$B\equiv (0,4)$$ whereas it should be $$B\equiv (4,0)$$. It does not change the answer for the radius but it certainly give a wrong answer for the center point.

Kind Regards,
Sudharaka.