A circle has a center that falls on the line y = 3/7x +1 and passes through ( 2 ,1 ) and (3 ,5 ). What is the equation of the circle?

1 Answer
May 17, 2016

(x-147/38)^2+(y -101/38)^2=(sqrt[4505/2]/19)^2

Explanation:

The circle equation is
C->(x-x_c)^2+(y-y_c)^2=r^2.
the straight y = 3/7x+1 can be written as
-3x+7(y-1)=0 or (p-p_0).vec v=0 where
p = (x,y), p_0=(0,1) and vec v = (-3,7)

The parametric representation for the straight line
is given by p = p_0 + lambda vec v^T where vec v^T is the vector with components (7,3) orthogonal to vec v. The circle center given by p_c=(x_c,y_c) is equidistant from p_1=(2,1) and p_2 = (3,5)

So we can equate
norm(p_c-p_1) = norm (p_c-p_2) but p_c = p_0+lambda_c vec v^T so we can state:
(p_0+lambda_c vec v^T-p_1).(p_0+lambda_c vec v^T-p_1)=(p_0+lambda_c vec v^T-p_1).(p_0+lambda_c vec v^T-p_1).
Developing and grouping
p_1.p_1-2p_0.p_1-2lambda_c vec v^T.p_1 = p_2.p_2-2p_0.p_2-2lambda_c vec v^T.p_2
or
p_2.p_2-p_1.p_1-2p_0.(p_2-p_1)-2lambda_c vec v^T.(p_2-p_1)=0
and finally
lambda_c = (p_2.p_2-p_1.p_1-2p_0.(p_2-p_1))/(2 vec v^T.(p_2-p_1))
Substituting values we obtain lambda_c = 21/38 then p_c = (147/38,101/38)