A circle has a center that falls on the line y = 3/2x +6 and passes through (1 ,2 ) and (6 ,1 ). What is the equation of the circle?

1 Answer
Narad T.
Oct 29, 2016

The equation of the circle is (x-44/7)^2+(y-108/7)^2=208.3

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, (1-a)^2+(2-b)^2=r^2
and (6-a)^2+(1-b)^2=r^2

Equating the equations
(1-a)^2+(2-b)^2=(6-a)^2+(1-b)^2
developing,
1-2a+a^2+4-4b+b^2=36-12a+a^2+1-2b+b^2

5-2a-4b=37-12a-2b
10a-2b=32
5a-b=16 this is equation 1
Also, y=(3x)/2+6
So b=(3a)/2+6
Solving for a and be, we get a=44/7 and b=108/7
We calculate the radius r^2=(6-44/7)^2+(1-108/7)^2

r^2=(2/7)^2+(108/7)^2=> r=14.43

Equation of circle is
(x-44/7)^2+(y-108/7)^2=208.3

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