the length of tangent from two points a b to a circle are 4,3 respectively .if a b are conjugate points then ab =

Dear Student,
let eqn of circle be x^2+y^2=r^2
length of tangent from a(x1,y1)is sqrt(x1^2+y1^2-r^2)=4
=>x1^2+y1^2-r^2=16 –(1)
similarly for b(x2,y2) we get x2^2+y2^2-r^2=9 –(2)
Now polar of (x1,y1) wrt circle:
xx1+yy1=r^2
as a and b are conjugate b satisfies this eqn.
so (x2,y2)=>x2x1+y2y1=r^2 –(3)
solving (1),(2) and (3),
we get a= (sqrt(r^2+16) , 0)
and b=(r^2/sqrt(r^2+16) , ± sqrt((25r^2+144)/(r^2+16))
[Ans]