If the line $y = mx + 1$ meets the circle $x^2 + y^2 + 3x = 0$ at two points equidistant from the $y$-axis and on opposite sides,then:

Let $P_1, P_2, P_3, P_4, P_5$ be five equally spaced points on the circumference of a circle of radius $1$,centred at $O$. Let $R$ be the set of points in the plane of the circle that are closer to $O$ than any of $P_1, P_2, P_3, P_4, P_5$. Then,$R$ is a

$A$ 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:

The length of the intercept that the circle $x^2 + y^2 + 10x - 6y + 9 = 0$ makes on the $x$-axis is:

If $\theta$ is the angle between the tangents drawn from the point $P(-1, -1)$ to the circle $x^2 + y^2 - 4x - 6y + c = 0$ and $\cos \theta = -\frac{7}{25}$,then the radius of the circle is

Let a triangle $ABC$ be inscribed in the circle $x^{2} - \sqrt{2}(x+y) + y^{2} = 0$ such that $\angle BAC = \frac{\pi}{2}$. If the length of side $AB$ is $\sqrt{2}$,then the area of the $\triangle ABC$ is equal to