The line $2x - 3y = 1$ divides the circular region $x^2 + y^2 \leq 6$ into two parts. If $S = \left\{ \left(2, \frac{3}{4}\right), \left(\frac{5}{2}, \frac{3}{4}\right), \left(\frac{1}{4}, -\frac{1}{4}\right), \left(\frac{1}{8}, \frac{1}{4}\right) \right\}$,then the number of points in the set $S$ that lie inside the smaller part is:

Suppose that two chords, drawn from the point $(1, 2)$ on the circle $x^2 + y^2 + x - 3y = 0$, are bisected by the $y$-axis. If the other ends of these chords are $R$ and $S$, and the midpoint of the line segment $RS$ is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:

Number of common tangents to the circles $x^2+y^2-6x-14y+48=0$ and $x^2+y^2-6x=0$ are

$A$ straight line through the vertex $P$ of a triangle $PQR$ intersects the side $QR$ at the point $S$ and the circumcircle of the triangle $PQR$ at the point $T$. If $S$ is not the centre of the circumcircle,then:
$(A) \frac{1}{PS}+\frac{1}{ST}<\frac{2}{\sqrt{QS \times SR}}$
$(B) \frac{1}{PS}+\frac{1}{ST}>\frac{2}{\sqrt{QS \times SR}}$
$(C) \frac{1}{PS}+\frac{1}{ST}<\frac{4}{QR}$
$(D) \frac{1}{PS}+\frac{1}{ST}>\frac{4}{QR}$

The two circles $x^2 + y^2 - 2x + 6y + 6 = 0$ and $x^2 + y^2 - 5x + 6y + 15 = 0$:

If the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ touches the $x$-axis,then: