A Circle S passes through the points of intersection of the circles x^2+y^2-2 x+2 y-2=0 and x^2+y^2+2 x-2 y+1=0 . If the centre of this circle S lies on the line x-y+6=0 , then the radius of the circle S is

Step-by-step solution

Since equation of circles is x^2+y^2-2 x+2 y-2+ aligned & k (x^2+y^2+2 x-2 y+1 )=0 & (1+k) x^2+(1+k) y^2+(2 k-2) x+(2-2 k) y+k-2=0 aligned x^2+y^2+ 2(k-1) k+1 x+ 2(1-k) 1+k y+ k-2 k+1 =0 So, centre = ( -(k-1) k+1 , -(1-k) 1+k )= ( 1-k 1+k , k-1 k+1 ) Since, centre lies on the line x-y+6=0 1-k 1+k - (k-1) 1+k +6=0 1-k-k+1 1+k =-6 aligned & 2-2 k=-6(1+k) & 1-k=-3-3 k k=-2 aligned So, radius = ( 1-k 1+k )^2+ ( k-1 k+1 )^2- ( k-2 k+1 ) = (-3)^2+(3)^2-4 = 14