Question 480006: the circle passes through the points (0, 0), (O, 5), (3, 3). Find the equation of the circle.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The circle passes through the points (0, 0), (O, 5), (3, 3). Find the equation of the circle.
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Standard form of equation for circle: (x-h)^2+(y-k)^2=r^2, (h,k) being the center of the circle with radius=r.
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3 equations from coordinates of given 3 points (0, 0), (0, 5), (3,3)
1) (0-h)^2+(0-k)^2=r^2
2) (0-h)^2+(5-k)^2=r^2
3) (3-h)^2+(3-k)^2=r^2
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solving for k
1) (0-h)^2+(0-k)^2=r^2
2) (0-h)^2+(5-k)^2=r^2
subtract eq2 fm eq1
k^2-(5-k)^2=0
k^2-(25-10k+k^2)=0
k^2-25+10k-k^2=0
25-10k=0
10k=25
k=2.5 or 5/2
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solving for h
2) (0-h)^2+(5-k)^2=r^2
3) (3-h)^2+(3-k)^2=r^2
subtract eq3 fm eq2
h^2+(5-2.5)^2-[(3-h)^2+(3-2.5)^2]=0
h^2+(2.5)^2-[9-6h+h^2+.5^2]=0
h^2+6.25-9+6h-h^2-.25=0
6.25-9.25+6h=0
-3=-6h
h=.5 or 1/2
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solving for r
1) (0-h)^2+(0-k)^2=r^2
(1/2)^2+(5/2)^2=r^2
1/4+25/4=r^2=26/4=13/2
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Equation of circle:
(x-1/2)^2+(y-5/2)^2=13/2
see graph below as visual check on answer
note: a more precise algebriac check can be made by plugging in coordinates of given points into equation
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y=±(13/2-(x-.5)^2)^.5+2.5
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