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)#