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

Question

Christian Antunez · Accepted Answer

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#

Sophie Adams · Answer

undefined To find the equation of a circle passing through three given points, you can use the general form of the equation of a circle, which is $ (x - h)^2 + (y - k)^2 = r^2 $. Substitute the coordinates of each point into this equation, which will result in three equations. Then, solve the system of equations to find the values of $ h $, $ k $, and $ r $, which represent the coordinates of the center of the circle and its radius, respectively. Once you have found $ h $, $ k $, and $ r $, substitute them back into the general equation of a circle to obtain the equation of the circle.