If three distinct points A, B, C are given in the 2dimensional coordinate plane such that the ratio of the distance of each one of them from the point (1,0) to the distance from (-1,0) is equal to 1 2 , then the circumcentre of the triangle A B C is at the point

Step-by-step solution

Let P(1,0) and Q(-1,0), A(x, y) aligned & Given: A P A Q = B P B Q = C P C Q = 1 2 & 2 A P=A Q & 4(A P)^2=A Q^2 aligned aligned & 4 [(x-1)^2+y^2 ]=(x+1)^2+y^2 & 4 (x^2+1-2 x )+4 y^2=x^2+1+2 x+y^2 & 3 x^2+3 y^2-8 x-2 x+4-1=0 & 3 x^2+3 y^2-10 x+3=0 & x^2+y^2- 10 3 x+1=0 aligned A lies on the circle given by (1). As B and C also follow the same condition. Centre of circumcircle of A B C= centre of circle given by (1) = ( 5 3 , 0 ) .