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

1 Answer
Jul 27, 2016

Answer:

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

Explanation:

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 #