A circle has a center that falls on the line #y = 2/7x +3 # and passes through # ( 2 ,4 )# and #(6 ,5 )#. What is the equation of the circle?

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Answer 1 · 2017-06-19T12:26:59

The equation of the circle is #(x-49/12)^2+(y-25/6)^2=629/144#

Let #C# be the mid point of #A=(2,4)# and #B=(6,5)#

#C=((2+6)/2,(4+5)/2)=(4,9/2)#

The slope of #AB# is #=(5-4)/(6-2)=1/4#

The slope of the line perpendicular to #AB# is #=-4#

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

#y-9/2=-4(x-4)#

#y=-4x+16+9/2=-4x+41/2#

The intersection of this line with the line #y=2/7x+3# gives the center of the circle.

#2/7x+3=-4x+41/2#

#2/7x+4x=41/2-3#

#30/7x=35/2#

#x=35/2*7/30=245/60=49/12#

#y=2/7*245/60+3=25/6#

The center of the circle is #(49/12,25/6)#

The radius of the circle is

#r^2=(49/12-2)^2+(25/6-4)^2#

#=(25/12)^2+(1/6)^2#

#=629/144#

The equation of the circle is

#(x-49/12)^2+(y-25/6)^2=629/144#
graph{((x-49/12)^2+(y-25/6)^2-629/144)(y-2/7x-3)=0 [-3.146, 9.34, 0.674, 6.914]}