How do you find the equation of the circle passing through p1(-3,6)p2(-5,2),p3(3,-6)?

1 Answer
May 26, 2016

Answer:

#(x-2)^2+(y-1)^2=( 5 sqrt(2))^2#

Explanation:

The circle equation is
#f(x,y)=(x-x_0)^2+(y-y_0)^2-r^2# in which #x_0,y_0# are the circle coordinates and #r# the radius.

If the points #p_1=(x_1,y_1), p_2=(x_2,y_2), p_3=(x_3,y_3)#
obey the circle equation then

#f(x_1,y_1) =f(x_2,y_2)=f(x_3,y_3)=0#

or
#{((x_1-x_0)^2+(y_1-y_0)^2-r^2 = 0),((x_2-x_0)^2+(y_2-y_0)^2-r^2 = 0),((x_3-x_0)^2+(y_3-y_0)^2-r^2 = 0):}#

Subtracting the second from the first and the third from the second we have

#{(2(x_2-x_1)x_0 + 2(y_2-y_1)y_0 +(x_1^2-x_2^2+y_1^2-y_2^2)=0),(2(x_3-x_2)x_0 + 2(y_3-y_2)y_0 +(x_2^2-x_3^2+y_2^2-y_3^2)=0):}#

Solving for #x_0,y_0# after substituting the values for #x_1,y_1,x_2,x_2,x_3,y_3# we get
#x_0=2,y_0=1#.

The radius is found by solving for #r#
#(x_1-x_0)^2+(y_1-y_0)^2-r^2 = 0# so #r = 5 sqrt(2)#