How do you write an equation for a circle passing through (-1,2), (3,4), and (2,-1)?

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Answer 1 · 2016-07-27T18:43:43

$(x-5/3)^2+(y-5/3)^2-65/9=0$

The circle's general equation is

$C->(x-x_0)^2+(y-y_0)^2-r^2=0$

This equation must be obeyed by the three points. Substituting ${x,y}$ for each point we have

${ ( 2 x_0 + x_0^2 - 4 y_0 + y_0^2+5 - r^2=0), ( - 6 x_0 + x_0^2 - 8 y_0 + y_0^2+25 - r^2=0), ( - 4 x_0 + x_0^2 + 2 y_0 + y_0^2+5 - r^2=0) :}$

Those equations are easily handled taking the difference between the first and the second, and the first and the third, obtaining

${ (2 x_0 + y_0-5=0), (x_0 - y_0=0) :}$

Solving for $x_0,y_0$ we obtain

${x_0=5/3, y_0=5/3}$

$r$ is obtained putting those results in any of the equations already obtained

$r =sqrt[65]/3$