If the point $(\lambda, 1+\lambda)$ lies inside the circle $x^2+y^2=1$,then

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

$A$ point $P$ is taken outside $\Delta ABC$ where $B(1, \sqrt{3})$,$A(0, 0)$,and $C(2, 0)$,but inside the acute angle $BAC$,such that $\angle APC = \frac{\pi}{6}$ and $\angle BPA = \frac{\pi}{12}$. The slope of the line $BP$ is:

If $(3,-2)$ is the centre of the circle $S \equiv x^2+y^2+2gx+2fy-23=0$ and $A$ is a point on the circle $S=0$ such that its distance from a point $P(-1,-5)$ is least,then $A=$

The triangle $PQR$ is inscribed in the circle $x^2 + y^2 = 25$. If $Q$ and $R$ have coordinates $(3, 4)$ and $(-4, 3)$ respectively,then $\angle QPR$ is equal to

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?