Question 125172: This problem is from a practice exam:
A circle has its center somewhere along the line y=3. It passes through the points (8,11) and (-4,-5). Find its equation.
-I think the first step is for me to find the radius. Because I know the equation of a circle is  . But I don't really know how to find the radius when I don't know the exact center.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A circle has its center somewhere along the line y=3. It passes through the points (8,11) and (-4,-5). Find its equation.
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Find the equation of the line thru the two points:
slope = (11--5)/(8--4) = 16/12 = 4/3
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The form is y = mx+b and you know x,y, and m; so solve for "b".
11 = (4/3)*8 +b
11 = (32/3) + b
b = 1/3
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EQUATION:
y = (4/3)x + (1/3)
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EQUATION for the Circle:
Let the center be (h,3)
(x-h)^2 + (y-k)^2 = r^2
Using point (8,11) to get:
(8-h)^2 + (11-3)^2 = r^2
Using point (-4,-5) to get:
(-4-h)^2 + (-5-3)^2 = r^2
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Simplify those equations to solve for h and r:
(64-16h+h^2) + 64 = r^2
(16+8h+h^2)+ 64 = r^2
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64-16h = 16+8h
h=0
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So the center of the circle is at (0,3)
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radius = sqrt[(8-0)^2+(11-3)^2] = sqrt[64 + 64] = = 8sqrt(2)
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EQUATION of the circle:
x^2 + (y-3)^2 = 128
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Cheers,
Stan H.
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