Prove that the centres of the circles x^2+y^2-10x+9=0,x^2+y^2-6x+2y+1=0 and x^2+y^2-18x-4y+21=0 lies on a line; find the equation of line on which they lie?

1 Answer
Feb 10, 2018

Centre of the 1st circle can be found by rearranging it,

given, # x^2 +y^2 -10 x +9=0#

or, #(x-5)^2 +(y-0)^2 -16=0#

So,its centre lies at #(5,0)#

for the 2nd one, #x^2 +y^2-6x+2y+1=0#

or, #(x-3)^2 +(y+1)^2 -9=0#

So,its centre lies at, #(3,-1)#

Similiarly for the 3rd one, #x^2+y^2-18x-4y+21=0#

or,#(x-9)^2 + (y-2)^2 -64=0#

So,its centre lies at #(9,2)#

So,the equation of a straight line passing through #(5,0) & (3,-1)# is expressed as,

#(y-0)/(x-5) = (y-(-1))/(x-3)#

or, #2y=x-5#

Now, putting #x=9# in this equation we get, #y=2#,that means this staright line passes through the point #(9,2)# (centre of the 3rd circle)

So,all three points lie on a starigh line,and the equation is stated above.