The equation of a line inclined at an angle $\frac{\pi}{4}$ to the $X$-axis,such that the two circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x - 14y + 65 = 0$ intercept equal lengths on it,is

$A$ line drawn from a fixed point $P(\alpha, \beta)$ intersects the circle $x^2 + y^2 = r^2$ at points $A$ and $B$. Then $PA \cdot PB = \dots$

If the length of the tangents drawn from the point $(1, 2)$ to the circles $x^2 + y^2 + x + y - 4 = 0$ and $3x^2 + 3y^2 - x - y + k = 0$ are in the ratio $4 : 3$,then $k =$

Let $L_1$ be a line passing through the origin and $L_2$ be the line $x + y = 1$. If the intercepts made by the circle $x^{2} + y^{2} - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations represents $L_1$?

The circle $C_1: x^2+y^2=3$,with centre at $O$,intersects the parabola $x^2=2y$ at the point $P$ in the first quadrant. Let the tangent to the circle $C_1$ at $P$ touch two other circles $C_2$ and $C_3$ at $R_2$ and $R_3$,respectively. Suppose $C_2$ and $C_3$ have equal radii $2\sqrt{3}$ and centres $Q_2$ and $Q_3$,respectively. If $Q_2$ and $Q_3$ lie on the $y$-axis,then:
$(A)$ $Q_2Q_3=12$
$(B)$ $R_2R_3=4\sqrt{6}$
$(C)$ Area of the triangle $OR_2R_3$ is $6\sqrt{2}$
$(D)$ Area of the triangle $PQ_2Q_3$ is $4\sqrt{2}$

Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then,which one of the following points lies on the circle $C$?