The angle between the circles $x^2+y^2-2x-9=0$ and $x^2+y^2-4y-1=0$ at their point of intersection is

  • A
    $\frac{\pi}{6}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{2}$

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For what values of $m$ does the circle $x^2 + y^2 = 4x + 8y + 5$ intersect the line $3x - 4y = m$ at two distinct points?

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$A$ circle $C$ of radius $1$ is inscribed in an equilateral triangle $PQR$. The points of contact of $C$ with the sides $PQ, QR, RP$ are $D, E, F$,respectively. The line $PQ$ is given by the equation $\sqrt{3}x + y - 6 = 0$ and the point $D$ is $\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$. Further,it is given that the origin and the centre of $C$ are on the same side of the line $PQ$.
$1.$ The equation of circle $C$ is
$(A) (x - 2\sqrt{3})^2 + (y - 1)^2 = 1$
$(B) (x - 2\sqrt{3})^2 + (y + \frac{1}{2})^2 = 1$
$(C) (x - \sqrt{3})^2 + (y + 1)^2 = 1$
$(D) (x - \sqrt{3})^2 + (y - 1)^2 = 1$
$2.$ Points $E$ and $F$ are given by
$(A) \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), (\sqrt{3}, 0)$
$(B) \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), (\sqrt{3}, 0)$
$(C) \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(D) \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$3.$ Equation of the sides $QR, RP$ are
$(A) y = \frac{2}{\sqrt{3}}x + 1, y = -\frac{2}{\sqrt{3}}x - 1$
$(B) y = \frac{1}{\sqrt{3}}x, y = 0$
$(C) y = \frac{\sqrt{3}}{2}x + 1, y = -\frac{\sqrt{3}}{2}x - 1$
$(D) y = \sqrt{3}x, y = 0$
Give the answer for questions $1, 2$ and $3$.

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If $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ are three given points,then the centre of the circle for which the lines $PQ, QR$ and $RP$ are the tangents is

The perimeter of a certain sector of a circle is equal to the length of the arc of the semicircle. Then,the angle at the centre of the sector in radians is

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