The condition that the lines joining the origin to the points of intersection of the line x a + y b =2 and the circle (x-a)^2+(y-b)^2=r^2 are at right angles is

Step-by-step solution

x a + y b =2 x 2 a + y 2 b =1 aligned & &(x-a)^2+(y-b)^2=r^2 & x^2+y^2-2 a x-2 b y+a^2+b^2-r^2=0 aligned Homogenizing eqn. (ii), we get aligned & x^2+y^2-2(a x+b y) 1+ (a^2+b^2-r^2 ) 1^2=0 & x^2+y^2-2(a x+b y) ( x 2 a + y 2 b )+ (a^2+b^2-r^2 ) & ( x 2 a + y 2 b )^2=0 & 4 a^2 b^2 (x^2+y^2 )-4 a b(a x+b y)(b x+a y)+ & (a^2+b^2-r^2 )(b x+a y)^2=0 & 4 a^2 b^2 (x^2+y^2 )-4 a b (a b x^2+a b y^2+ (a^2+b^2 ) x y )+ & (a^2+b^2-r^2 )(b x+a y)^2=0 & 4 a^2 b^2 (x^2+y^2 )-4 a^2 b^2 (x^2+y^2 )-4 a b & (a^2+b^2 ) x y+ (a^2+b^2-r^