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

1 Answer
Shwetank Mauria
Jul 28, 2017

#36x^2+36y^2-396x-732y+1095=0#

Explanation:

As ordinates of points #(8,2)# and #(3,2)# are same, the line joining them is parallel to #x#-axis and its equation is #y=2# and the midpoint of points #(8,2)# and #(3,2)# being #(11/2,2)#, center lies on #x=11/2#.

Further center lies on the pont of intersection of #y=5/3x+1# and #x=11/2# and therefore we have

#y=5/3xx11/2+1=55/6+1=61/6# and center is #(11/2,61/6)# and as circle passes through #(3,2)# the radius is

#sqrt((3-11/2)^2+(2-61/6)^2)=sqrt(25/4+2401/36)=sqrt(2626/36)#

and hence equation of circle is

#(x-11/2)^2+(y-61/6)^2=2626/36#

or #x^2-11x+121/4+y^2-61/3y+3721/36=2626/36#

or #36x^2+36y^2-396x-732y+2184=0#

or #9x^2+9y^2-99x-183y+546=0#

graph{(9x^2+9y^2-99x-183y+546)((x-8)^2+(y-2)^2-0.1)((x-3)^2+(y-2)^2-0.1)(2x-11)((x-11/2)^2+(y-61/6)^2-0.1)(3y-5x-3)=0 [-16, 24, 0, 20]}

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