Let $a=1+i$ and $z=x+i y$. If the curve $z \bar{z}+a z+\bar{a} \bar{z}-4=0$ is cut by the straight line $(z+\bar{z})-i(z-\bar{z})+2=0$ at two points $A$ and $B$, then the equation of the circle passing through the origin, $A$ and $B$ is

A point $P$ moves so that distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p, 0

If $P A$ and $P B$ are the tangents drawn from the point $P(1,1)$ to the circle $x^2+y^2+g x+g y-2=0$ with $C$ as the centre, then the area (in sq. units) of the quadrilateral $P A C B$ is