A variable triangle `A B C` is circumscribed about a fixed circle of unit radius. Side `B C` always touches the circle at D and has fixed direction. If B and C vary in such a way that (BD) (CD)=2, then locus of vertex A will be a straight line. parallel to side BC perpendicular to side BC making an angle `(pi/6)` with BC making an angle `sin^(-1)(2/3)` with `B C`

Options

A. parallel to side BC
B. perpendicular to side BC
C. making an angle `(pi//6)` with BC
D. making an angle `sin^(-1) (2//3)` with BC

Answer

A

Explanation

`BD = (s -b), CD = (s-c)`
`rArr (s-b) (s-c) =2`
or `s(s-a) (s-b) (s-c) = 2 s(s -a)`
or `Delta^(2) = 2s (s-a)`
or `(Dleta^(2))/(s^(2)) = (2(s-a))/(s)` (using `Delta = rs`)
or `r^(2) = (2(s-a))/(s)`
or `(a)/(s)` = constant
Now, `Delta = (1)/(2) aH_(a)`, where `H_(a)` is the distance of A from BC. Thus,
`(Delta)/(2) = (1)/(2) (aH_(a))/(s) = 1 " or " H_(a) = (2s)/(a)` = constant
Therefore, locus of A will be a straight line parallel to side BC