Circle

Here if we add center of circles G₁, G₂, G₃ ....... G_n, then we get a polygon of n sides.

From figure you can see one side of polygon makes angle $\theta$ with the center.

$\therefore$ n sides make angle = n$\theta$

We know, $n\theta = 2\pi $

$ \Rightarrow \theta = {{2\pi } \over n}$

Here triangle OMN is an isosceles triangle. Line joining of point O and midpoint O of MN (point A) is perpendicular to line MN and perpendicular bisector of angle $\theta$.

$\therefore$ $\angle MOA = {\theta \over 2} = {\pi \over n}$

From right angle triangle OMA,

$\sin {\pi \over n} = {r \over {R + r}}$

$ \Rightarrow R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over n}$

Option A:

When $n = 4$,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 4}$

$ \Rightarrow R + r = \sqrt 2 r$

$ \Rightarrow R = \left( {\sqrt 2 - 1} \right)r$

$\therefore$ Option (A) is wrong.

Option B :

When $n = 5$,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 5} = r{\mathop{\rm cosec}\nolimits} 36^\circ $

We know, in 0 to ${\pi \over 2}$, ${\mathop{\rm cosec}\nolimits} \theta $ is decreasing.

$\therefore$ ${\mathop{\rm cosec}\nolimits} 45^\circ < {\mathop{\rm cosec}\nolimits} 36^\circ < {\mathop{\rm cosec}\nolimits} 30^\circ $

$ \Rightarrow \sqrt 2 < {\mathop{\rm cosec}\nolimits} 36^\circ < 2$

$ \Rightarrow \sqrt 2 r < r{\mathop{\rm cosec}\nolimits} 36^\circ < 2r$

$ \Rightarrow \sqrt 2 r - r < r{\mathop{\rm cosec}\nolimits} 36^\circ - r < 2r - r$

$ \Rightarrow \left( {\sqrt 2 - 1} \right)r < R < r$

$\therefore$ $R < r$, when $n = 5$

$\therefore$ Option (B) is wrong.

Option C :

When $n = 8$,

$R + r = r{\mathop{\rm cosec}\nolimits} {\pi \over 8}$

${\mathop{\rm cosec}\nolimits} \theta$ is decreasing function in 0 to ${\pi \over 2}$.

$\therefore$ ${\mathop{\rm cosec}\nolimits} {\pi \over 8} > {\mathop{\rm cosec}\nolimits} {\pi \over 4}$

$ \Rightarrow r{\mathop{\rm cosec}\nolimits} {\pi \over 8} > \sqrt 2 r$

$ \Rightarrow R + r > \sqrt 2 r$

$ \Rightarrow R > \left( {\sqrt 2 - 1} \right)r$

$\therefore$ Option (C) is correct.

Option D :

When $n = 12$, then

$R + r = r\left( {\cos ec{\pi \over {12}}} \right)$

$ \Rightarrow R + r = r \times {{2\sqrt 2 } \over {\sqrt 3 - 1}}$ [as $\cos ec{\pi \over {12}} = {{2\sqrt 2 } \over {\sqrt 3 - 1}}$]

$ \Rightarrow R + r = r \times {{2\sqrt 2 \left( {\sqrt 3 + 1} \right)} \over 2}$

$ \Rightarrow R + r = \sqrt 2 r\left( {\sqrt 3 + 1} \right)$

$ \Rightarrow R = \left[ {\sqrt 2 \left( {\sqrt 3 + 1} \right) - 1} \right]r$

$ \Rightarrow R < \sqrt 2 \left( {\sqrt 3 + 1} \right)r$

$\therefore$ Option (D) is correct.

Let, P $\equiv$ (cos$\theta$, sin$\theta$)

$\therefore$ equation of tangent and normal at P

x cos$\theta$ + y sin$\theta$ = 1 ..... (1)

and y = x tan$\theta$ ...... (2)

Now, equation of tangent at S : x = 1 ...... (3)

Solving (1) and (3), Q $\equiv$ (1, cosec$\theta$ $-$ cot$\theta$)

$\therefore$ equation of straight line parallel to RS drawn from Q

y = cosec$\theta$ $-$ cot$\theta$ ..... (4)

Let, E $\equiv$ (h, k)

$\therefore$ k = h tan$\theta$ [from (2)]

or, $\tan \theta = {k \over h}$

Again, k = cosec$\theta$ $-$ cot$\theta$ [from (4)]

or, $k = {{1 - \cos \theta } \over {\sin \theta }}$

or, $k = {{1 - {h \over {\sqrt {{h^2} + {k^2}} }}} \over {{k \over {\sqrt {{h^2} + {k^2}} }}}} = {{\sqrt {{h^2} + {k^2}} - h} \over k}$

or, ${k^2} = \sqrt {{h^2} + {k^2}} - h$

or, $h + {k^2} = \sqrt {{h^2} + {k^2}} $

$\therefore$ locus of E $x + {y^2} = \sqrt {{x^2} + {y^2}} $

Clearly, points $\left( {{1 \over 3},{1 \over {\sqrt 3 }}} \right)$ and $\left( {{1 \over 3}, - {1 \over {\sqrt 3 }}} \right)$ are on locus of E.

Therefore, (A) and (C) are the correct options.

Let, the equation of the required circle is

${x^2} + {y^2} + 2gx + 2fy + c = 0$ ..... (1)

Circle (I) cuts the circle ${(x - 1)^2} + {y^2} = 16$

i.e., ${x^2} + {y^2} - 2x = 15$ orthogonally

$\therefore$ $2( - g + 0) = - 15 + c$

or, $ - 2g = - 15 + c$

The circle (1) also cuts the circle ${x^2} + {y^2} = 1$ orthogonally.

$\therefore$ 0 = $-$1 + c or, c = 1

$\therefore$ g = 7

Now, the circle (1) passes through the point (0, 1).

$\therefore$ $2f + 1 + c = 0$ or, $2f + 1 + 1 = 0$ or, f = $-$1

$\therefore$ the equation of the required circle is

${x^2} + {y^2} + 14x - 2y + 1 = 0$

whose centre is ($-$7, 1) and radius $ = \sqrt {49 + 1 - 1} = 7$ units

Therefore, (B) and (C) are the correct option.

Note :

The condition of the circle ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ cuts orthogonally to the circle ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ is $2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$

Let a circle $x^2+y^2+2 g x+2 f y+c=0$ touches $x$-axis and have $y$-intercept of length $2 \sqrt{7}$.

$\begin{aligned} & \therefore 2 \sqrt{g^2-c}=0 \text { and } 2 \sqrt{f^2-c}=2 \sqrt{7} \\ & \Rightarrow g^2=c \text { and } f^2-c=7 \\ & \Rightarrow c=g^2=f^2-7 \quad \text{... (i)} \end{aligned}$

Given, the circle touches $x$-axis at a distance 3 unit from origin.

$\begin{aligned} & \therefore|-g|=3 \\ & \Rightarrow g= \pm 3 \end{aligned}$

Put $g= \pm 3$ in the equation

$\Rightarrow c=9 \text { and } f= \pm 4$

Hence, the possible equation of circles are

$\begin{aligned} & x^2+y^2-6 x+8 y+9=0, \\ & x^2+y^2-6 x-8 y+9=0, \end{aligned}$

$\begin{aligned} & x^2+y^2+6 x+8 y+9=0 \text { and } x^2+y^2+6 x-8 y \\ & +9=0 \end{aligned}$

Hints:

The length of X-intercept and Y-intercept of the circle $x^2+y^2+2 g x+2 f y+c=0$ are $2 \sqrt{g^2-c}$ and $2 \sqrt{f^2-c}$ respectively.