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

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Answer 1 · 2017-01-18T14:03:13

#(x - 9/2)^2 + (y - 10)^2 = 153/4#

Circle C: #(x - a)^2 + (y - b)^2 = R^2#

#(a,b) in r: b = 2/3 a + 7#

#C: (x - a)^2 + (y - 2/3 a - 7)^2 = R^2#

#(3, 4) in C: (3 - a)^2 + (4 - 2/3 a - 7)^2 = R^2# // equation 1

#(6, 4) in C: (6 - a)^2 + (4 - 2/3 a - 7)^2 = R^2# // equation 2

Subtract: (1) - (2)

#(3 - a)^2 - (6 - a)^2 = 0#

#3 - a = 6 - a#   or   #3 - a = - (6 - a)#

#0a = 3#   or   #3 + 6 = a + a#

#a = 9/2#

#b = 2/3 * 9/2 + 7 = 10#

(1) #\Rightarrow R^2 = (3 - 9/2)^2 + (4 - 10)^2#

#R^2 = (3/2)^2 + 36 = (9 + 144)/4 = 153/4#