MHB Equation of a Circle: Center on y=6-2x, Passing Through (-2,0) & (4,0)

AI Thread Summary
The discussion focuses on finding the equation of a circle with its center on the line y=6-2x, passing through points A(-2,0) and B(4,0). The center is identified as (x₀, 6-2x₀), leading to the equation of the circle in Cartesian coordinates. By substituting the points A and B into this equation, two equations are derived to solve for the unknowns x₀ and the radius r. The correct center is determined to be C(3,0) with a radius of 5, confirmed through calculations involving the midpoint and perpendicular bisector of line segment AB. The discussion concludes with clarification on the coordinates of point B to ensure accuracy in the solution.
thorpelizts
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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?
 
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thorpelizts said:
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.
 
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 said:
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.
 
soroban said:
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.
 
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