A circle has a center that falls on the line #y = 2x +4 # and passes through #(4 ,4 )# and #(1 ,2 )#. What is the equation of the circle?

1 Answer
Konstantinos Michailidis
Mar 28, 2016

Hence the circle has a center on #y=2x+4# its coordinates are
#(a,2a+4)#

Now the distance of the two points from the center is equal hence

#sqrt[(4-a)^2+(4-2a-4)^2]=sqrt[(1-a)^2+(2-2a-4)^2]=> [(4-a)^2+(4-2a-4)^2]=[(1-a)^2+(2-2a-4)^2]=> a=11/14#

Hence the center is #(11/14,39/7)# and the radius is

#r=sqrt[(4-11/14)^2+(4-39/7)^2]=sqrt(2509)/14#

Hence the equation of the circle becomes

#(x-11/14)^2+(y-39/7)^2=(sqrt(2509)/14)^2#

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