Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is $$2y \over x^2$$. If the curve passes through the centre of the circle x² + y² – 2x – 2y = 0, then its equation is :

The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y+ 4z = 5 which is perpendicular to the plane x – y + z = 0 is :

The tangent to the parabola y² = 4x at the point where it intersects the circle x² + y² = 5 in the first quadrant, passes through the point :

Let ƒ : R $$ \to $$ R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0.
Then $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$$ is equal to