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 :

Let $$C$$ be the centre of the circle $$x^{2}+y^{2}-x+2 y=\frac{11}{4}$$ and $$P$$ be a point on the circle. A line passes through the point $$\mathrm{C}$$, makes an angle of $$\frac{\pi}{4}$$ with the line $$\mathrm{CP}$$ and intersects the circle at the points $$Q$$ and $$R$$. Then the area of the triangle $$P Q R$$ (in unit $$^{2}$$ ) is :

A circle $$C_{1}$$ passes through the origin $$\mathrm{O}$$ and has diameter 4 on the positive $$x$$-axis. The line $$y=2 x$$ gives a chord $$\mathrm{OA}$$ of circle $$\mathrm{C}_{1}$$. Let $$\mathrm{C}_{2}$$ be the circle with $$\mathrm{OA}$$ as a diameter. If the tangent to $$\mathrm{C}_{2}$$ at the point $$\mathrm{A}$$ meets the $$x$$-axis at $$\mathrm{P}$$ and $$y$$-axis at $$\mathrm{Q}$$, then $$\mathrm{QA}: \mathrm{AP}$$ is equal to :

If the circle $$x^{2}+y^{2}-2 g x+6 y-19 c=0, g, c \in \mathbb{R}$$ passes through the point $$(6,1)$$ and its centre lies on the line $$x-2 c y=8$$, then the length of intercept made by the circle on $$x$$-axis is :