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.
#1
thorpelizts
6
0
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).
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.
#3
soroban
191
0
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).
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?
#4
Sudharaka
Gold Member
MHB
1,558
1
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,
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.
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