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

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Answer 1 · 2016-08-21T10:00:07

#(x-4)^2#+ #(y-11/3)^2# =#34/9#

The general equation of a circle is
#(x-a)^2# + #(y-b)^2#=#r^2#
Where (a,b) is the centre of the circle and r is the radius.

So (a,b) is on the line y=#2/3# x +1
Substituting b=#2/3#a+1. Equation 1

(5,2) is on the circle so
#(5-a)^2# +#(2-b)^2#=#r^2#. Equation 2

(3,2) is on the circle so
#(3-a)^2#+#(2-b)^2# =#r^2# Equation 3

Subtract equation 3 from equation 2 gives
#(5-a)^2# -#(3-a)^2# =0

Multiply out and simplify gives a= 4

Substitute in equation 1 gives b= #11/3#

Substitute in equation 2 gives #r^2# =#34/9#

Put all values into equation 3 as a check.

Yes it is correct