A circle through the common points of the circles x^2+y^2-2x-4y+1=0 and x^2+y^2+2x-6y+1=0 has the centre on the line 4y-7x-19=0.find the centre and radius of the circle?

1 Answer
Feb 21, 2018

Contd.......

Explanation:

The eqn. of the circle #S# that passes through the points of

intersection of the circles #S_1 and S_2#, where,

# S_1 : x^2+y^2+2g_1x+2f_1y+c_1=0,and, #

# S_2 : x^2+y^2+2g_2x+2f_2y+c_2=0,# is given by,

# S : S_1+lambda*S_2=0, i.e., #

# S : (x^2+y^2+2g_1x+2f_1y+c_1)+lambda(x^2+y^2+2g_2x+2f_2y+c_2)=0, lambda in RR-{-1}.#

Accordingly, let, the reqd. circle be,

#S:(x^2+y^2-2x-4y+1)+lambda(x^2+y^2+2x-6y+1)=0#.

#:.S:(1+lambda)x^2+(1+lambda)y^2+2(lambda-1)x+2(-3lambda-2)y+(lambda+1)=0, or, #

#S: x^2+y^2+2((lambda-1)/(lambda+1))x+2((-3lambda-2)/(lambda+1))y+1=0#.

The centre #C" of "S" is "C(-((lambda-1)/(lambda+1)),-((-3lambda-2)/(lambda+1)))#.

Given that, #C" lies on the line : "4y-7x-19=0#, we have,

#4((3lambda+2)/(lambda+1))+7((lambda-1)/(lambda+1))=19#.

#:. 12lambda+8+7lambda-7=19lambda+19, i.e., #

#19lambda