JEE Math Practice Question

Let y = y(x) be a function of x satisfying $$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}} $$ where k is a constant and $$y\left( {{1 \over 2}} \right) = - {1 \over 4}$$. Then $${{dy} \over {dx}}$$ at x = $${1 \over 2}$$, is equal to :

1. A.$${2 \over {\sqrt 5 }}$$
2. B.$$ - {{\sqrt 5 } \over 2}$$
3. C.$${{\sqrt 5 } \over 2}$$
4. D.$$ - {{\sqrt 5 } \over 4}$$