A circle has a center that falls on the line #y = 7/4x +4 # and passes through # ( 3 ,7 )# and #(7 ,5 )#. What is the equation of the circle?

1 Answer
Narad T.
Oct 26, 2016

The equation of the circle is #(x-32)^2+(y-60)^2=3650#

Explanation:

Let the center of the circle be #(a,b)# and radius r
Then the equation of the circle is
#(x-a)^2+(y-b)^2=r^2#

Then, #(3-a)^2+(7-b)^2=r^2#
and #(7-a)^2+(5-b)^2=r^2#

Equqting the equations
#(3-a)^2+(7-b)^2=(7-a)^2+(5-b)^2#
developing,
#9-6a+a^2+49-14b+b^2=49-14a+a^2+25-10b+b^2#

#9-6a-14b=25-14a-10b#
#8a-4b=16#
#2a-b=4# this is equation 1
Also, #y=(7x)/4+4#
So #b=(7a)/4+4#
Solving for a and be, we get #a=32# and #b=60#
We calculate the radius #r^2=(7-32)^2+(5-60)^2#

#r^2=25^2+55^2##=># #r=60.41#

Equation of circle is
#(x-32)^2+(y-60)^2=3650#

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