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

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Answer 1 · 2017-05-27T05:45:14

The equation of the circle is #(x-21/5)^2+(y-8)^2=1261/25#

Let the points #A=(8,2)# and #B=(3,1)#

The midpoint of #AB# is #C=((8+3)/2, (1+2)/2)=(11/2,3/2)#

The slope of #AB# is #m=(1-2)/(3-8)=-1/-5=1/5#

The slope of the line perpendicular to #AB# is #m'=-5#, as

#mm'=-1#

The equation of the line, passing through #C# and perpendicular to #AB# is #L'#

#y-3/2=-5(x-11/2)#

#y-3/2=-5x+55/2#

#y=-5x+55/2+3/2=-5x+29#

Let the line #y=5/3x+1# be #L#

The center of the circle #O# lies at the intersection of #L# and #L'#

We have 2 equations

#y=5/3x+1# and #y=-5x+29#

So,

#5/3x+1=-5x+29#

#5/3x+5x=29-1=28#

#20/3x=28#

#x=28/20*3=21/5#

#y=5/3*21/5+1=8#

The center of the circle is #O=(21/5,8)#

The radius of the circle is

#r^2=(21/5-3)^2+(8-1)^2=36/25+49=1261/25#

The equation of the circle is

#(x-21/5)^2+(y-8)^2=1261/25# graph{(y-5/3x-1)(y+5x-29)((x-21/5)^2+(y-8)^2-1261/25)=0 [-18.56, 21.97, -2.56, 17.72]}