How do you find the equation of a circle that passes through points (-8,-2)(1-,-11) and (-5,9)?

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Answer 1 · 2018-06-01T11:47:10

The equn. of a circle is :

# 7x^2+7y^2-52x-10y-912=0#

Let the equation of the circle be

#x^2+y^2+2gx+2fy+c=0,#

Since the circle passes through #(-8,-2), (1,-11)# and #(-5,9)#

#(-8)^2+(-2)^2-16g-4f+c=0#
#1^2+(-11)^2+2g-22f+c=0#
#(-5)^2+9^2-10g+18f+c=0#

These equations simplify to

#-16g-4f+c+68=0# #-----------(1)#
#2g-22f+c+122=0# #-----------(2)#
#-10g+18f+c+106=0# #----------(3)#

Subtracting #(1)# from #(2)#
#18g-18f+54=0# #------------(4)#

Subtracting #(1)# from #(3)#
#6g+22f+38=0# #-------------(5)#

Multiply #(5)# by 3
#18g+66f+114=0# #------------(6)#

Subtracting #(4)# from #(6)#
#84f+60=0#

#therefore f= -5/7#

Substituting the value of #f# into #(4)#

#18g+90/7+54=0#
#18g+468/7=0#
#18g=-468/7#

Divide both sides by 18
#g=-26/7#

Substituting the value of f and g into #(1)#
#-16(-26/7)-4(-5/7)+c+68=0#
#416/7+20/7+c+68=0#
#c+912/7=0#
#c=-912/7#

Hence, the equation of the circle is

#x^2+y^2+2(-26/7)x+2(-5/7)y-912/7=0#
#x^2+y^2-52/7x-10/7y-912/7=0#

Multiply through by 7
#therefore 7x^2+7y^2-52x-10y-912=0#

In the form
#(x-a)^2+(y-b)^2=r^2#

#(x-26/7)^2+(y-5/7)^2=7085/49#