A circle has a center that falls on the line #y = 3x +4 # and passes through #(4 ,4 )# and #(1 ,2 )#. What is the equation of the circle?

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Answer 1 · 2017-06-24T12:15:59

The equation of the circle is #(x-11/18)^2+(y-35/6)^2=2405/162#

Let #C# be the mid point of #A=(4,4)# and #B=(1,2)#

#C=((4+1)/2,(4+2)/2)=(5/2,3)#

The slope of #AB# is #=(2-4)/(1-4)=(-2)/(-3)=2/3#

The slope of the line perpendicular to #AB# is #=-3/2#

The equation of the line passing trrough #C# and perpendicular to #AB# is

#y-3=-3/2(x-5/2)#

#y=-3/2x+15/4+3=-3/2x+27/4#

The intersection of this line with the line #y=3x+4# gives the center of the circle.

#3x+4=-3/2x+27/4#

#3x+3/2x=27/4-4#

#9/2x=11/4#

#x=11/18#

#y=3*(11/18)+4=35/6#

The center of the circle is #(11/18,35/6)#

The radius of the circle is

#r^2=(1-11/18)^2+(2-35/6)^2#

#=(7/18)^2+(-23/6)^2#

#=4810/324=2405/162#

The equation of the circle is

#(x-11/18)^2+(y-35/6)^2=2405/162#
graph{((x-11/18)^2+(y-35/6)^2-2405/162)(y-3x-4)(y+3/2x-27/4)=0 [-9.25, 10.75, -2.56, 7.44]}