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coordinate geometry

thorpelizts

New member
Sep 7, 2012
6
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
Feb 5, 2012
1,621
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
Feb 2, 2012
409
Hello, thorpelizts!

Another approach . . .


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

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

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

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

Their intersection (and hence the center) is: [tex]C(3,0).[/tex]

The radius is: [tex]AC = BC = 5.[/tex]

Got it?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 5, 2012
1,621
Hello, thorpelizts!

Another approach . . .



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

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

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

Their intersection (and hence the center) is: [tex]C(3,0).[/tex]

The radius is: [tex]AC = BC = 5.[/tex]

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.