A circle has a center that falls on the line #y = 2/3x +7 # and passes through # ( 3 ,1 )# and #(6 ,4 )#. What is the equation of the circle?

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Answer 1 · 2016-05-14T12:17:07

The equation of the circle is #x^2+y^2-14y+4=0#

As the center will be equidistant from #(3,1)# and #(6,4)#, the center will lie on the line

#(x-3)^2+(y-1)^2=(x-6)^2+(y-4)^2# or

#x^2-6x+9+y^2-2y+1=x^2-12x+36+y^2-8y+16#

#12x-6x+8y-2y+10-52=0# or #6x+6y-42=0# or

#x+y=7# but as it also lies on #y=2/3x+7#, putting this in first

#x+2/3x+7=7# or #x=0# and hence #y=7#. Hence center is #(0,7)#.

Its distance from #(3,1)# is #sqrt((0-3)^2+(7-1)^2)=sqrt(9+36)=sqrt45#, i.e radius

Hence equation of circle is

#(x-0)^2+(y-7)^2=(sqrt45)^2#

#x^2+y^2-14y+49=45# or #x^2+y^2-14y+4=0#

graph{(x^2+y^2-14y+4)(y-(2/3)x-7)(x+y-7)=0 [-17.5, 22.5, -4.24, 15.76]}