A circle has a center that falls on the line #y = 5/8x +1 # and passes through # ( 5 ,2 )# and #(3 ,4 )#. What is the equation of the circle?

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Answer 1 · 2017-06-10T11:59:42

The equation of the circle is #(x-16/3)^2+(y-13/3)^2=50/9#

Let #C# be the mid point of #A=(5,2)# and #B=(3,4)#

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

The slope of #AB# is #=(4-2)/(3-5)=-2/2=-1#

The slope of the line perpendicular to #AB# is #=1#

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

#y-3=1(x-4)#

#y=x-1#

The intersection of this line with the line #y=5/8x+1# gives the center of the circle.

#5/8x+1=x-1#

#3/8x=2#

#x=16/3#

#y=16/3-1=13/3#

The center of the circle is #(16/3,13/3)#

The radius of the circle is

#r^2=(16/3-5)^2+(13/3-2)^2#

#=(1/3)^2+(7/3)^2#

#=50/9#

The equation of the circle is

#(x-16/3)^2+(y-13/3)^2=50/9#

graph{((x-16/3)^2+(y-13/3)^2-50/9)(y-5/8x-1)(y-x+1)=0 [0.604, 9.372, 1.62, 6.002]}