How do you write the equation for a circle that passes through the points at (4,5), (-2,3) and (-4,-3)?

1 Answer
Jun 26, 2016

Answer:

#x^2+y^2-6x+4y-37=0#

Explanation:

let's substitute each coordinate of all points in the general equation of the circle:

#x^2+y^2+ax+by+c=0#

so that you have three equations with unknown variables a,b,c:

#4^2+5^2+4a+5b+c=0# and
#(-2)^2+3^2-2a+3b+c# and
#(-4)^2+(-3)^2-4a-3b+c=0#

let's solve in a few steps:
#4a+5b+c+41=0 and -2a+3b+c+13=0 and -4a-3b+c+25=0#

#c=-4a-5b-41 and #(by substituting c) #-2a+3b-4a-5b-41+13=0 and -4a-3b-4a-5b-41+25=0#

#-6a-2b-28=0 and -8a-8b-16=0#

that is equivalent to:

#3a+b+14=0 and a+b+2=0#

By subtracting you obtain:

#2a+12=0#

#a=-6#

so, by substituting the obtained value of a:

#-6+b+2=0#

#b=4#

so

#c=-4(-6)-5(4)-41=24-20-41=-37#

The equation of the circle is:

#x^2+y^2-6x+4y-37=0#