A circle has a center that falls on the line #y = 7/6x +1 # and passes through #(1 ,4 )# and #(8 ,5 )#. What is the equation of the circle?

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Answer 1 · 2016-07-16T03:24:10

#(7x-30)^2 +49(y-6)^2 = 725#

The equation of the circle is

#(x-x_c)^2 +(y-y_c)^2 = R^2 #

where #(x_c,y_c)# is the center and #R# the radius of the circle.

To find the center note that since the points #(1,4)# and #(8,5)# are on the circle, we have

#(1-x_c)^2 +(4-y_c)^2 = (8-x_c)^2 +(5-y_c)^2 = R^2 #

Rearranging, we get a linear equation obeyed by #(x_c,y_c)#

#7x_c+y_c=36#

However, we already know

#y_c = 7/6 x_c +1#

Solving this pair of linear simultaneous equations gives

#x = 30/7, y=6#

Substituting this back in the equation #(1-x_c)^2 +(4-y_c)^2 = R^2 #, we get #R^2 = 725/49#.