How do you write the equation of the circle containing the points J(-6, 0), K(-3, 3), and L(0, 0)?

1 Answer
Mar 19, 2018

Answer:

The equation of circle is #x^2+y^2+6x= 0#

Explanation:

Let the equation of circle be #x^2+y^2+2gx+2fy +c= 0#

Points #j(-6,0), k(-3,3),l(0,0)#

#:. (-6)^2+0^2+2g*(-6)+2f*0 +c= 0 #

#:. (36-12g +c= 0 or -12g+36+c=0(1) #, similarly

#:. (-3)^2+3^2+2g*(-3)+2f*3 +c= 0 #

#:. (9+9-6g +6f +c= 0 or -6g+6f+18+c=0(2) #

and #:. 0^2+0^2+2g*0+2f*0 +c= 0 #

#:. c= 0# . Equation (1) is now #-12g+36+0=0 #

#:. 12g=36 or g=3#.Equation (2) is now

#-6g+6f+18+c=0 or -6*3+6f+18+0=0 # or

#-18+6f+18=0 or 6f=0 :. f=0# . Hence the equation of

circle is #x^2+y^2+6x= 0#

graph{x^2+y^2+6x=0 [-10, 10, -5, 5]}