A circle has a center that falls on the line #y = 7/9x +7 # and passes through # ( 7 ,3 )# and #(5 ,1 )#. What is the equation of the circle?

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Answer 1 · 2016-10-24T10:43:14

THe equation of the circle is #(x-11/16)^2+(y-117/16)^2=58.4#

Let the center of the circle be #a,b#

Then the equation of the circle is
#(x-a)^2+(y-b)^2=r^2#
Putting the points #(7,3)# and #(5,1)# in the equation of the circle

#(7-a)^2+(3-b)^2=r^2# and #(5-a)^2+(1-b)^2=r^2#
so #(7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2#
Expanding the equations

#49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2#

#49-14a+9-6b=25-10a+1-2b#

#58-14a-6b=26-10a-2b#
#32=4a+4b# #=># #8=a+b# this is equation 1

From the equation of the line, we have #b=(7a)/9+1#
#9b=7a+9# this is equation 2
Solving for a and b , we get
#a=11/16# and #b=117/16#

Then we calculate #r#
#r^2=(7-11/16)^2+(3-117/16)^2#
#r^2=(101^2+69^2)/16#
#r=7.65#
So the equation of the circle is
#(x-11/16)^2+(y-117/16)^2=58.4#