If the pair of lines joining the origin and the points of intersection of the line a x + b y = 1 and the curve x 2 + y 2 - x - y - 1 = 0 are at right angles, then the locus of the point ( a , b ) is a circle of radius

Step-by-step solution

The equation of the line is given as a   x + b   y = 1 and the equation of the curve is given as, x 2 + y 2 - x - y - 1 = 0 By homogenising both equations, x 2 + y 2 - x ( a x + b y ) - y ( a x + b y ) - ( a x + b y ) 2 = 0 x 2 + y 2 - a x 2 - b x y - a x y - b y 2 - a 2 x 2 + b 2 y 2 + 2 a b x y = 0 x 2 1 - a - a 2 + x y ( - a - b - 2 a b ) + y 2 1 - b - b 2 = 0 Since it is given that the pair of lines joining the origin and the points of intersection of the line and curve are at right angles. Coefficien