A circle has a center that falls on the line #y = 3/8x +3 # and passes through # ( 1 ,4 )# and #(2 ,9 )#. What is the equation of the circle?

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Answer 1 · 2017-06-22T10:12:01

The equation of the circle is #(x-152/23)^2+(y-126/23)^2=17797/529#

Let #C# be the mid point of #A=(1,4)# and #B=(2,9)#

#C=((1+2)/2,(9+4)/2)=(3/2,13/2)#

The slope of #AB# is #=(9-4)/(2-1)=(5)/(1)=5#

The slope of the line perpendicular to #AB# is #=-1/5#

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

#y-13/2=-1/5(x-3/2)#

#y=-1/5x+3/10+13/2=-1/5x+34/5#

The intersection of this line with the line #y=3/8x+3# gives the center of the circle.

#3/8x+3=-1/5x+34/5#

#3/8x+1/5x=34/5-3#

#23/40x=19/5#

#x=19/5*40/23=152/23#

#y=3/8*152/23+3=126/23#

The center of the circle is #(152/23,126/23)#

The radius of the circle is

#r^2=(1-152/23)^2+(4-126/23)^2#

#=(129/23)^2+(-34/23)^2#

#=17797/529#

The equation of the circle is

#(x-152/23)^2+(y-126/23)^2=17797/529#
graph{((x-152/23)^2+(y-126/23)^2-17797/529)(y-3/8x-3)=0 [-8.25, 20.23, -2.34, 11.9]}