- #1
thorpelizts
- 6
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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?
poor in circles. how to even start?
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?
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]
\|
6*
|\
| \
| \
(0,4)oB \
/| \
/ | \
* | \
/ * \
A / | * \3
- - o - - + - - * - + - - -
(-2,0) | * \
| oC
| \ *
|
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.
soroban said:Hello, thorpelizts!
Another approach . . .
The center lies on the line [tex]y \,=\,6-2x[/tex]Code:\| 6* |\ | \ | \ (0,4)oB \ /| \ / | \ * | \ / * \ A / | * \3 - - o - - + - - * - + - - - (-2,0) | * \ | oC | \ * |
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?
The equation of a circle with center on y=6-2x and passing through (-2,0) and (4,0) can be found by using the formula (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center point and r is the radius. In this case, the center point is (2,6) and the radius is 2. Therefore, the equation is (x-2)^2 + (y-6)^2 = 4.
To determine the center point of a circle given its equation, you can rearrange the equation into the form (x-h)^2 + (y-k)^2 = r^2 and then the center point will be (h,k).
Yes, the center point of a circle can be on a line. In the case of this specific equation, the center point is on the line y=6-2x.
A circle with center on y=6-2x and passing through (-2,0) and (4,0) will have two points, (-2,0) and (4,0). These points are the intersections of the circle with the x-axis.
The radius of a circle can be found by taking the square root of the number on the right side of the equation in the form (x-h)^2 + (y-k)^2 = r^2. In the case of this specific equation, the radius is 2.