A circle has a center that falls on the line #y = 1/7x +4 # and passes through # ( 7 ,8 )# and #(3 ,1 )#. What is the equation of the circle?

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Answer 1 · 2016-12-27T22:29:19

The equation of a circle with center #(x_o,y_o)# and radius #R# is

#(x-x_o)^2+(y-y_o)^2=R^2#

Hence the center belongs to the line we have that

#y_o=1/7*x_o +4#

Because the circle passes from point #(7,8)# and #(3,1)# we have also

#(7-x_o)^2+(8-y_o)^2=R^2#

and

#(3-x_o)^2+(1-y_o)^2=R^2#

Hence the RHS of both equations is the same (#=R^2#)
we have

#(7-x_o)^2+(8-y_o)^2=(3-x_o)^2+(1-y_o)^2#

#7^2-14*x_o + (x_o)^2+8^2-16*y_o + (y_o)^2= 3^2-6*x_o + (x_o)^2+1-2*y_o + (y_o)^2#

#49+64-14*x_o-16*y_o-9+6*x_o-1+2*y_o=0#

#-8*x_o-14*y_o + 103=0#

#8*x_o + 14*y_o - 103=0#

Because #y_o=1/7*x_o +4# and #8*x_o + 14*y_o - 103=0#

solving the system of these two equation yields the center of the

circle

#(x_o,y_o)=(47/10,327/70)#

and the radius is

#(3-47/10)^2+(1-327/70)^2=R^2#

#R=sqrt(8021/490)#

Finally the equation of the circle is

#(x-47/10)^2+(y-327/70)^2=8021/490#