JEE Math Practice Question

Let y = y(x) be a solution of the differential equation, $$\sqrt {1 - {x^2}} {{dy} \over {dx}} + \sqrt {1 - {y^2}} = 0$$, |x| < 1. If $$y\left( {{1 \over 2}} \right) = {{\sqrt 3 } \over 2}$$, then $$y\left( { - {1 \over {\sqrt 2 }}} \right)$$ is equal to :

1. A.$$ - {{\sqrt 3 } \over 2}$$
2. B.None of those
3. C.$${{1 \over {\sqrt 2 }}}$$
4. D.$$-{{1 \over {\sqrt 2 }}}$$