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

1 Answer
Narad T.
Oct 24, 2016

The equation of the circle is (x-27/16)^2+(y-101/16)^2=39.2

Explanation:

Let the center of the circle be (a,b)
Then the equation of the circle is (x-a)^2+(y-b)^2=r^2
As the circles passes through (7,3) and (5,1)
Then substituting those points in the equation of the circle

(7-a)^2+(3-b)^2=r^2
(5-a)^2+(1-b)^2=r^2

so (7-a)^2+(3-b)^2=(5-a)^2+(1-b)^2
Developing both sides
49-14a+a^2+9-6b+b^2=25-10a+a^2+1-2b+b^2
simplifying
58-14a-6b=26-10a-2b
4a+4b=32
a+b=8 this is equation 1
Putting (a,b) in the equation of the line
b=7/9a+5 this is equation 2
Solving for a and b
we obtain (27/16,101/16) as the center of the circle
The radius r=sqrt (85^2+63^2)/16=6.26
The equation of the circle is (x-27/16)^2+(y-101/16)^2=39.2

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