The points of intersection of the line $$ax + by = 0,(a \ne b)$$ and the circle $${x^2} + {y^2} - 2x = 0$$ are $$A(\alpha ,0)$$ and $$B(1,\beta )$$. The image of the circle with AB as a diameter in the line $$x + y + 2 = 0$$ is :

The locus of the mid points of the chords of the circle $${C_1}:{(x - 4)^2} + {(y - 5)^2} = 4$$ which subtend an angle $${\theta _i}$$ at the centre of the circle $$C_1$$, is a circle of radius $$r_i$$. If $${\theta _1} = {\pi \over 3},{\theta _3} = {{2\pi } \over 3}$$ and $$r_1^2 = r_2^2 + r_3^2$$, then $${\theta _2}$$ is equal to :

Let the tangents at two points $$\mathrm{A}$$ and $$\mathrm{B}$$ on the circle $$x^{2}+\mathrm{y}^{2}-4 x+3=0$$ meet at origin $$\mathrm{O}(0,0)$$. Then the area of the triangle $$\mathrm{OAB}$$ is :

For $$\mathrm{t} \in(0,2 \pi)$$, if $$\mathrm{ABC}$$ is an equilateral triangle with vertices $$\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t)$$ and $$C(a, b)$$ such that its orthocentre lies on a circle with centre $$\left(1, \frac{1}{3}\right)$$, then $$\left(a^{2}-b^{2}\right)$$ is equal to :