A circle has a center that falls on the line #y = 5/2x +1 # and passes through #(1 ,2 )# and #(6 ,1 )#. What is the equation of the circle?

1 Answer
Jan 12, 2017

#5x^2+5y^2-68x-180y+403=0#

Explanation:

let center of circle is #(x,y)# and r = radius of circle.

at point #(1,2)#
#(x-1)^2+(y-2)^2=r^2#
#(x-1)^2+(5/2x+1-2)^2=r^2#
#(x-1)^2+(5/2x-1)^2=r^2# #rArr#a

at point #(6,1)#
#(x-6)^2+(y-1)^2=r^2#
#(x-6)^2+(5/2x+1-1)^2=r^2#
#(x-6)^2+(5/2x)^2=r^2##rArr#b

from a&b, #a=b#

#(x-1)^2+(5/2x-1)^2=(x-6)^2+(5/2x)^2#
#cancelx^2-2x+1+cancel(25/4x^2)-5x+1=cancelx^2-12x+36+cancel(25/4x^2)#
#-2x+1-5x+1=-12x+36#
#-7x+12x=36-2#
#5x=34#,#x=34/5#

#y=5/2(34/5)+1#
#y=18#
so the center of circle is #(34/5,18)#

Therefore,the equation of circle
#(x-34/5)^2+(y-18)^2=(6-34/5)^2+(1-18)^2#
#(x-34/5)^2+(y-18)^2=(-4/5)^2+(-17)^2#
#(x-34/5)^2+(y-18)^2=16/25+289#
#(x-34/5)^2+(y-18)^2=7241/25#
#(5x-34)^2/25+(y-18)^2=7241/25#

multiply by 25
#(5x-34)^2+25(y-18)^2=7241#
#25x^2-340x+1156+25y^2-900y+8100-7241=0#
#25x^2+25y^2-340x-900y+1156+8100-7241=0#
#25x^2+25y^2-340x-900y+2015=0#

divided by 5, so the equation of circle is
#5x^2+5y^2-68x-180y+403=0#