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

1 Answer
Narad T.
Nov 1, 2016

The equation of the circle is #(x+40/13)^2+(y+18/13)^2=10.51^2#

Explanation:

Let #(a,b)# be the center of the circle
As the center lies on the line, we have #b=(7a)/4+4#
Then #(x-a)^2+(y-b)^2=r^2#, and #r# is the radius of the circle
So, #(3-a)^2+(7-b)^2=r^2#
and #(7-a)^2+(1-b)^2=r^2#
#:.# #(3-a)^2+(7-b)^2=(7-a)^2+(1-b)^2#
#9-6a+a^2+49-14b+b^2=49-14a+a^2+1-2b+b^2#
#58-6a-14b=50-14a-2b#
#12b=8a+8##=>##3b=2a+2#
Solving for #(a,b)# with the two equations
We find #a=-40/13# and #b=-18/13#

Then #r=sqrt((7+(40/13))^2+(1+(18/13))^2)#
#r=10.51#
Having #r#, #a#, and #b#, we write the equation of the circle

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