A circle has a center that falls on the line #y = 5/6x +8 # and passes through #(4 ,8 )# and #(2 ,5 )#. What is the equation of the circle?

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Answer 1 · 2017-05-12T11:24:21

The equation of the circle is #(x-1/3)^2+(y-149/18)^2=4381/324#

Let the line #L# be #6y-5x=48#

Let the points #A# and #B# be

#A=(4,8)#

#B=(2,5)#

The mid-point of #AB# is

#C=((4+2)/2,(8+5)/2)=(3,13/2)#

The slope of #AB# is #m=(5-8)/(2-4)=3/2#

The slope of the line perpendicular to #AB# is

#m'=-2/3# as #mm'=-1#

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

#y-13/2=-2/3(x-3)#

#6y-39=-4x+12#

#6y+4x=51# This is line #L'#

We need the point of intersection of line #L# and #L'#

#6y-5x=48#

#6y+4x=51#

#-9x=-3#

#x=1/3#

#6y=5/3+48=149/3#

#y=149/18#

The center of the circle is #O=(1/3,149/18)#

The radius of the circle is

#r^2=((2-1/3)^2+(5-149/18)^2)#

#=25/9+3481/324#

#=4381/324#

The equation of the circle is

#(x-1/3)^2+(y-149/18)^2=4381/324#

graph{(y-5/6x-8)((x-1/3)^2+(y-149/18)^2-4381/324)=0 [-8.29, 7.505, 4.45, 12.35]}