let the center be (a,2).
SO radius is sqrt(a^2 + 2^2) = sqrt(a^2+4) [sqrt = square root]
or, radius is also sqrt[(a-1)^2 + (2-1)^2] = sqrt[ (a-1)^2 + 1]
so from the above two lines, sqrt(a^2+4) = sqrt[ (a-1)^2 + 1]
or, a^2 + 4 = a^2 - 2a + 1 +1
or, a = -1
so the center is at (-1,2) and radius is sqrt(5)
so the eqn is (x + 1)^2 + (y - 2)^2 = 5
2007-06-25 21:40:53
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answer #1
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answered by Mock Turtle 6
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Let (a,b) be the center and r be the radius.
Thus, the equation of the circle is:
(x - a)² + (y - b)² = r²
Since the center passes through y = 2, then b = 2.
The center is now (a,2)
Then, the equation of the circle is:
(x - a)² + (y - 2)² = r²
Now, since it passes through (0,0) and (1,1), then
(0 - a)² + (0 - 2)² = r²
and
(1 - a)² + (1 - 2)² = r²
We simplify both equations
a² + 4 = r²
1 - 2a + a² + 1 = r²
then
a² + 4 = r²
a² - 2a + 2 = r²
Then, we equate
a² + 4 = a² - 2a + 2
Then we solve for a
2a = 2 - 4
2a = -2
a = -1
Therefore, we can now solve for r²
a² + 4 = r²
r² = 1 + 4
r² = 5
Therefore, the equation of the circle is
(x + 1)² + (y - 2)² = 5
That equation is in the center-radius form.
Or, if you want it to be in general form, then we expand
x² + 2x + 1 + y² - 4y + 4 = 5
Therefore the equation in general form is:
x² + 2x + y² - 4y = 0
2007-06-26 05:39:59
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answer #2
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answered by kevin! 5
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the circle has its center on the line y=2 , lets suppose the x-co-ordinate of the center is x.
so the center of the circle is the point (x,2)
Distance from the center of the circle to any point on the circle is the radius. So distance between (x,2) and (0,0) is radius. So is the distance between (x,2) and (1,1)
Distance between (x,2) and (0,0) = Sqrt (x^2 +4)
Distance between (x,2) and (1,1)= sqrt [(x-1)^2 +1]
Since both of the above are same: we can equate and solve for x::
x^2 +4 =+ or - [(x-1)^2 +1]...since both the sides are squared
taking + on the right hand side: we get x = -1
taking - on the right hand side : we get x as an imaginary number.
So the center of the circle is the point : (-1,2)
Equation of any circle is : (x-a)^2 + (y-b)^2 = r^2
where (a,b) is the center of the circle and r is the radius of the circle.
Substituting all the values: we get
(x+1)^2 + (y-2)^2 = 5
2007-06-26 05:10:16
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answer #3
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answered by Surendra G 1
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The center of the circle is the intersection of the perpendicular bisector of the line segment between (0,0) and (1,1) and the line y = 2.
The line thru (0,0) and (1,1) is y = x. Its slope is 1.
The slope of the perpendicular bisector is the negative reciprocal or m = -1. It passes thru the midpoint of the two given points, namely thru (1/2, 1/2).
The equation of the perpendicular bisector is:
y - 1/2 = -1(x - 1/2)
y - 1/2 = -x + 1/2
y = -x + 1
Find the intersection of this line with the line y = 2. Set the equations equal.
y = -x + 1 = 2
-x = 1
x = -1
The center of the circle is the point (-1, 2).
The radius is the distance from (-1, 2) to (0, 0).
r² = (-1-0)² + (2-0)² = 1 + 4 = 5
The equation of the circle is:
(x + 1)² + (y - 2)² = 5
2007-06-26 04:50:34
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answer #4
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answered by Northstar 7
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The center would be at a point then (a,2) and the equation would be (x-a)^2 + (y-2)^2 = r^2.
(0,0) --> a^2 + 2^2 = a^2 + 4 = r^2
(1,1) --> (1-a)^2 + (-1)^2 = a^2 - 2a + 2 = r^2
a^2 + 4 = a^2 - 2a + 2 (since r^2 = r^2)
a = -1, pretty trivially
r^2 = a^2 + 4 = (-1)^2 + 4 = 5
(x + 1)^2 + (y - 2)^2 = 5
2007-06-26 05:09:11
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answer #5
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answered by jcsuperstar714 4
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Centre also lies on the perpendicular bisector of ( 0, 0 ) and ( 1, 1 ), the equation of which is
( x - 0 )^2 + ( y - 0 )^2 = ( x - 1 )^2 + ( y - 1 )^2
or, x + y = 1 ... ( 1 )
Solving it with y = 2 ... ( 2 ) gives the centre as C ( -1, 2 )
Radius of the circle is the distance between (0, 0) and (-1, 2)
Therefore, radius, r = ( -1 )^2 + ( 2 )^2 = sq. rt. 5
Hence, equation of the circle is
( x +1 )^2 + ( y - 2)^2 = 5
or, x^2 + y^2 + 2x - 4y = 0.
2007-06-26 05:50:39
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answer #6
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answered by Madhukar 7
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Two points on a circle are always equidistant from the center. With given information, the x-coordinates for the center of the circle has to be '-1' with radius sqrt (5).
Equation would be:
sq (x+1) + sq (y-2) = 5
Where sq stands for SQUARE
and sqrt stands for SQUARE ROOT
2007-06-26 04:42:49
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answer #7
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answered by go-green-save-earth 1
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(x - a)^2 + (y - 2)^2 = r^2
(0 - a)^2 + (0 - 2)^2 = r^2
a^2 + 4 = r^2
(1 - a)^2 + (1 - 2)^2 = r^2
2 - 2a + a^2 = r^2
a^2 + 4 = 2 - 2a + a^2
4 = 2 - 2a
2a = - 2
a = -1
r = â5
(x + 1)^2 + (y - 2)^2 = 5
1 + 4 = 5
4 + 1 = 5
2007-06-26 04:57:54
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answer #8
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answered by Helmut 7
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