Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

  • A
    $2+\sqrt{3}$
  • B
    $3+\sqrt{3}$
  • C
    $4$
  • D
    $2+\frac{1}{\sqrt{3}}$

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Let the centre of the circle be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is . . . . . .

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$A$ line meets the circle $x^2+y^2-4x-4y-8=0$ at two points $A$ and $B$. If $P(2,-2)$ is a point on the circle such that $PA=PB=2$,then the equation of the line $AB$ is:

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Consider the following statements:
$I$. The intercept made by the circle $x^2+y^2-2x-4y+1=0$ on $Y$-axis is $2\sqrt{3}$.
$II$. The intercept made by the circle $x^2+y^2-4x-2y+6=0$ on $X$-axis is $2\sqrt{2}$.
$III$. The straight line $y=2x+1$ cuts the circle $x^2+y^2=9$ at two distinct points.
Which one of the following options is correct?
$(a)$ $I$: True,$II$: True,$III$: True
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$(c)$ $I$: True,$II$: False,$III$: True
$(d)$ $I$: False,$II$: False,$III$: True

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