$A$ circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in the circle is

  • A
    $\frac{a^2}{3}$
  • B
    $\frac{2a^2}{3}$
  • C
    $\frac{a^2}{6}$
  • D
    $\frac{a^2}{12}$

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If from any point on the circle $x^2+y^2+2gx+2fy+c=0$,tangents are drawn to the circle $x^2+y^2+2gx+2fy+c \sin^2 \alpha + (g^2+f^2) \cos^2 \alpha = 0$,where $0 < \alpha < \frac{\pi}{2}$,then the angle between those tangents is

Let $M = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \leq r^2\}$,where $r > 0$. Consider the geometric progression $a_n = \frac{1}{2^{n-1}}$,$n = 1, 2, 3, \ldots$. Let $S_0 = 0$ and,for $n \geq 1$,let $S_n$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$,let $C_n$ denote the circle with center $(S_{n-1}, 0)$ and radius $a_n$,and $D_n$ denote the circle with center $(S_{n-1}, S_{n-1})$ and radius $a_n$.
$(1)$ Consider $M$ with $r = \frac{1025}{513}$. Let $k$ be the number of all those circles $C_n$ that are inside $M$. Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then
$(A)$ $k + 2l = 22$ $(B)$ $2k + l = 26$ $(C)$ $2k + 3l = 34$ $(D)$ $3k + 2l = 40$
$(2)$ Consider $M$ with $r = \frac{(2^{199}-1)\sqrt{2}}{2^{198}}$. The number of all those circles $D_n$ that are inside $M$ is
$(A)$ $198$ $(B)$ $199$ $(C)$ $200$ $(D)$ $201$

Let the line $L: \sqrt{2}x + y = \alpha$ pass through the point of intersection $P$ (in the first quadrant) of the circle $x^2 + y^2 = 3$ and the parabola $x^2 = 2y$. Let the line $L$ touch two circles $C_1$ and $C_2$ of equal radius $2\sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the $y$-axis,then the square of the area of the triangle $PQ_1Q_2$ is equal to:

If a circle of constant radius $3k$ passes through the origin and meets the axes at $A$ and $B$,the locus of the centroid of the triangle $OAB$ is the circle

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If the line $x \cos \theta + y \sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,then the value of $\theta$ is:

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