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 insuch a way that (BD) (CD)=2, then locus of vertex Awill be a straight line.parallel to side BCperpendicular to side BCmaking an angle `(pi/6)`with BCmaking an angle `sin^(-1)(2/3)`with `B C`

A variable triangle `A B C`is circumscribed about a fixed circle of un

1.	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 insuch a way that (BD) (CD)=2, then locus of vertex Awill be a straight line.parallel to side BCperpendicular to side BCmaking an angle `(pi/6)`with BCmaking an angle `sin^(-1)(2/3)`with `B C`
Answer» Here, `BD = s-b` `CD = s-c` `BDCD = (s-b)(s-c)` `=>2 = (s-b)(s-c)` `=>2s(s-a) = s(s-a)(s-b)(s-c)` `=>2s(s-a) = Delta^2` `=>Delta^2/s^2 = (2(s-a))/s` `=>r^2 = 2-(2a)/s` As, `r` is constant, so `a/s` will be constant. Let `H_a` is the distance of `A` from `BC`. Then, `Delta = 1/2a*H_a` `=>Delta/s = 1/(2s)aH_a` It is given that circle is of unit radius, ` :. 1 = a/(2s)H_a` `=>H_a = (2s)/a` It means `H_a` will be constant as `a/s` is a constant. So, locus of vertex `A` will be a straight line parallel to `BC`. So, option `(a)` is the correct option.