Question 771838: find the center and radius of the circle passes through (2,3), (6,1) and (4,-3).
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39618)   (Show Source):
You can put this solution on YOUR website!  fundamental standard form for circle
 ____________and use this form with your given points to obtain a system of three equations, and then solve the system; but this way may be messy. Your constants are the unknowns and they are in the form of complicated variable expressions.
Easier would be to take the typical general form for a conic section as 
 _________THIS is the form to start with, forming one equation each for each given point. Your unknowns will be the unknown coefficients d, e, f.
Solve THAT system, substitution, or elimination, or through matrix operations, or Cramer's Rule. Once that is done, you would write the resulting general form equation, Complete the Square to turn into standard form equation; and then directly read the center and radius.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! find the center and radius of the circle passes through (2,3), (6,1) and (4,-3).
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solve for h, k, and r^2 with following system of 3 equations:
..
(2,3) (2-h)^2+(3-k)^2=r^2
(6,1) (6-h)^2+(1-k)^2)=r^2
(4,-3) (4-h)^2+(-3-k)^2)=r^2
..
(2-h)^2+(3-k)^2=r^2
(6-h)^2+(1-k)^2=r^2
expand:
4-4h+h^2+9-6k+k^2=r^2
36-12h+h^2+1-2k+k^2=r^2
subtract:
-32+8h+8-4k=0
-24+8h-4k=0
-6+2h-k=0
..
(6-h)^2+(1-k^2)=r^2
(4-h)^2+(-3-k)^2=r^2
expand:
36-12h+h^2+1-2k+k^2=r^2
16-8h+h^2+9+6k+k^2=r^2
subtract:
20-4h-8-8k=0
12-4h-8k=0
3-h-2k=0
..
-6+2h-k=0
3-h-2k=0
..
-12+4h-2k=0
3-h-2k=0
subtract
-15+5h=0
5h=15
h=3
k=-6+2h=0
(2-h)^2+(3-k)^2=r^2
(2-3)^2+(3)^2=r^2
r^2=1+9=10
Equation of given circle:
center: (3,0)
radius: √10
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