Circle

$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$

$\because$ $\left( {1, - {1 \over 3}} \right)$ lies on or inside the C

then $4 + {4 \over 9} - 12 - {8 \over 3} + k \le 0$

$ \Rightarrow k \le {{92} \over 9}$

Now, circle lies in 4^th quadrant centre $ \equiv \left( {{3 \over 2}, - 1} \right)$

$\therefore$ $r < 1 \Rightarrow \sqrt {{9 \over 4} + 1 - {k \over 4}} < 1$

$ \Rightarrow {{13} \over 4} - {k \over 4} < 1$

$ \Rightarrow {k \over 4} > {9 \over 4}$

$ \Rightarrow k > 9$

$\therefore$ $k \in \left( {9,{{92} \over 9}} \right)$

Equation of perpendicular bisector of AB is

$y - {3 \over 2} = - {1 \over 5}\left( {x - {5 \over 2}} \right) \Rightarrow x + 5y = 10$

Solving it with equation of given circle,

${(x - 5)^2}{\left( {{{10 - x} \over 5} - 1} \right)^2} = {{13} \over 2}$

$ \Rightarrow {(x - 5)^2}\left( {1 + {1 \over {25}}} \right) = {{13} \over 2}$

$ \Rightarrow x - 5 = \pm \,{5 \over 2} \Rightarrow x = {5 \over 2}$ or ${{15} \over 2}$

But $x \ne {5 \over 2}$ because AB is not the diameter.

So, centre will be $\left( {{{15} \over 2},{1 \over 2}} \right)$

Now, ${r^2} = {\left( {{{15} \over 2} - 2} \right)^2} + {\left( {{1 \over 2} + 1} \right)^2}$

$ = {{65} \over 2}$

Let the centre be (h, k)

So, $\left| h \right| = \left| {{{h + k} \over {\sqrt 2 }}} \right|$

$ \Rightarrow 2{h^2} = {h^2} + {k^2} + 2hk$

Locus will be ${x^2} - {y^2} = 2xy$

Co-ordinate of centre $C \equiv \left( {{{3 + ( - 1)} \over 2},{{ - 6 - 2} \over 2}} \right) \equiv (1, - 2)$

L₁ is passing through A

$ \Rightarrow - 4 - 6 + {K_1} = 0$

$ \Rightarrow {K_1} = 10$

L₂ is passing through B

$ \Rightarrow 12 + 18 + {K_2} = 0$

$ \Rightarrow {K_2} = - 30$

Equation of ${L_1}:4x - 3y + 10 = 0$

Equation of ${L_2} = 4x - 3y - 30 = 0$

Diameter of circle $ = \left| {{{10 + 30} \over {\sqrt {{4^2} + {{( - 3)}^2}} }}} \right| = 8$

$\Rightarrow$ Radius = 4

Equation of circle ${(x - 1)^2} + {(y + 2)^2} = 16$

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.

Circumference is divided in $1: 2$.

So, $\angle A C B=\frac{360^{\circ}}{3}=120^{\circ}$

So, $\angle D C B=60^{\circ}$

$ P=\frac{|4(5)+3(3)-4|}{\sqrt{16+9}}=\frac{25}{5}=5 $

In $\triangle B C D, \cos 60^{\circ}=p / r$

$ \begin{array}{ll} \Rightarrow & \frac{1}{2}=\frac{p}{r} \\ \Rightarrow & r=2 p \\ & r=2 \times 5=10 \end{array} $

Equation of circle,

$ (x-5)^2+(y-3)^2=(10)^2 $

Coordinate of $D$ is $(-5,4)$

$O$ is Mid-point of $B D$.

Let coordinate of $B$ is $(a, b)$

$ \begin{array}{rlrlrl} & \Rightarrow & \frac{a-5}{2} & =3 \text { and } \frac{b+4}{2}=-4 \\ & \Rightarrow & a & =11 \\ & \text { and } & b & =-12 \end{array} $

∴ Coordinate of $B$ is $(11,-12)$.

As $A B \| D C$.

Equation of $A B$ is $y=-12$

Coordinate of $A$ is $(-5,-12)$

Equation of circle is

$ \begin{aligned} & (x-3)^2+(y+4)^2=(-5-3)^2+(4+4)^2 \\ \Rightarrow \quad & \quad(x-3)^2+(y+4)^2=128 \\ \Rightarrow & x^2+y^2-6 x+8 y-103=0 \end{aligned} $

Tangent at point ( $-5,4$ ) is

$ \begin{array}{ll} & -5 x+4 y-3(x-5)+4(y+4)-103=0 \\ \Rightarrow & -5 x+4 y-3 x+15+4 y+16-103=0 \\ \Rightarrow & -8 x+8 y-72=0 \\ \Rightarrow & x-y+9=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Tangent at point $(-5,-12)$

$ \begin{aligned} & \quad-5 x-12 y-3(x-5)+4(y-12)-103=0 \\ & \Rightarrow-5 x-12 y-3 x+15+4 y-48-103=0 \\ & \quad \begin{aligned}\Rightarrow -8 x-8 y-136 & =0 \\ x+y+17 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} \\ & \end{aligned} $

Adding Eqs. (i) and (ii),

$ \begin{aligned} 2 x+26 & =0 \\ x & =-13 \end{aligned} $

Substitute $x=-13$ in Eq.

$ \begin{aligned} -13-y+9 & =0 \\ y & =-4 \end{aligned} $

∴ The required point is $(-13,-4)$.

Given equation of circle

$ x^2+y^2=3 $

Chord of contacts of tangents drawn from $L_1, L_2$ and $L_3$ are $2 x=3, x-2 y=3$ and $4 x+4 y=3$ respectively.

For concurrent lines,

$ \begin{aligned}Let\,\, \Delta & =\left|\begin{array}{ccc} 2 & 0 & -3 \\ 1 & -2 & -3 \\ 4 & 4 & -3 \end{array}\right| \\ & =2(6+12)-3(4+8) \\ & =36-36 \\ & =0 \end{aligned} $

Hence, the three lines are concurrent.

Given,

$ x^2+y^2+2 x=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Centre of the given circle is $C_1(-1,0)$ and radius $\left(r_1\right)$ is 1 unit

$ x^2+y^2-2 y-3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Centre of the circle is $C_2(0,1)$ and radius $\left(r_2\right)$ is 2 units

$ \begin{aligned} C_1 C_2 & =\sqrt{(-1)^2+(1)^2}=\sqrt{2} \\ r_1+r_2 & =1+2=3 \text { units } \\ C_1 C_2 &

$ \begin{array}{rlrl} &A C_1 =1 \text { unit and } B C_2=2 \text { units } \\ \triangle P A C_1 & \sim \triangle P B C_2 \\ \therefore \quad & \frac{P A}{P B} =\frac{1}{2}=\frac{P C_1}{P C_2} \end{array} $

$\Rightarrow C_1 A$ is mid-point of $P C_2$. Let the coordinate of $P$ be $(h, k)$

$ \begin{aligned} & & \frac{h+0}{2} & =-1 \text { and } \frac{k+1}{2}=0 \\ \Rightarrow & & h & =-2 \text { and } k=-1 \end{aligned} $

∴ Coordinate of $P$ is $(-2,-1)$.

Equation of line passes through $(-2,-1)$ is

$ \begin{array}{lrl} & y+1=m(x+2) \\ \Rightarrow & m x-y+2 m-1=0 \\ & \text { As } P B \perp B C_2, & \\ & \frac{m \times 0-1+2 m-1}{\sqrt{1+m^2}}= \pm 2 & \\ \Rightarrow & \frac{2(m-1)}{\sqrt{1+m^2}}= \pm 2 & \\ \Rightarrow & \frac{m-1}{\sqrt{1+m^2}}= \pm 1 & \\ \Rightarrow & m^2-2 m+1=1+m^2 \\ \Rightarrow & 2 m=0 \Rightarrow m=0 \end{array} $

∴ Equation of $P A$ is $y=-1$

Slope of $C_1 C_2\left(m_1\right)=1$

$ \begin{aligned} \tan \theta & =\left|\frac{1-0}{1+1 \times 0}\right|=1 \\ \theta & =45^{\circ} \\ \angle A P M & =90^{\circ} \end{aligned} $

Let $m_2$ be the slope of line $P M$.

Equation of line $P M$ is

$ y+1=\frac{1}{0}(x+2) \Rightarrow x=-2 $

∴ The combined common tangent is

$ \begin{array}{r} (y+1)(x+2)=0 \\ \Rightarrow \quad x y+2 y+x+2=0 \end{array} $

Let the equation of circle whose centre is $(h, k)=$

$ \begin{aligned} & (x-h)^2+(y-k)^2=r^2 \\ \Rightarrow & x^2+h^2-2 x h+y^2+k^2-2 k y=r^2 \\ \Rightarrow & x^2+y^2-2 x h-2 k y=r^2-h^2-k^2 \\ \Rightarrow & x^2+y^2-2 x h-2 k y+c=0 \end{aligned} $

(Let $h^2+k^2-r^2=c$ )

Thus also pass origin then,

$ \begin{array}{cc} c=0 \\ \therefore \quad x^2+y^2-2 x h-2 k y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Eq. (i) is orthogonal to

$ x^2+y^2+2 x+6 y+12=0 $

and $x^2+y^2+4 x-6 y+9=0$

then $\quad-4 h+6 k=9\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

and $\quad-2 h-6 k=12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Adding Eqs. (ii) and (iv), we get

$ \begin{aligned} -6 h & =21 \\ h & =-\frac{21}{6}=\frac{-7}{2} \end{aligned} $

Putting the value of $h$ in Eq. (ii), we get

$ k=-\frac{5}{6} $

$ \begin{aligned}Now,\,\, k-2 h & =-\frac{5}{6}-\left(\frac{-7}{2}\right) \times 2 \\ & =\frac{-5+42}{6}=\frac{37}{6} \end{aligned} $

Let $S_1=x^2+y^2+2 g x-4 y+4=0$

$ \begin{aligned}& S_2=x^2+y^2+6 x+2 f y+12=0 \\and\,\, & S_3=x^2+y^2+10 y+20=0 \end{aligned} $

For radical centre, we have

$ \begin{aligned} & S_1=S_2=S_3 \\ \Rightarrow & S_1-S_2=0 \\ \Rightarrow & (2 g-6) x-(4+2 f) y-8=0 \end{aligned} $

Since, $(-1,-1)$ is the radical centre.

$ \begin{aligned} & \therefore(2 g-6)(-1)-(4+2 f)(-1)-8=0 \\ & \Rightarrow \quad g-f=1 \end{aligned} $

Centre lies on the line $x+y=5$.

Let centre be ( $x_1, 5-x_1$ )

Perpendicular distance from centre $C$ to $y-5=0$ Perpendicular distance from centre to line $x-2=0$

$ \left|5-x_1-2\right|=\left|x_1-5\right| $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 3-x_1= \pm\left(x_1-5\right) \\ & \text { Now } \quad 3-x_1=x_1-5 \\ & \Rightarrow x_1=4 \text { and } 3-x_1=-x_1+5 \\ & \Rightarrow \quad 3 \neq 5, x_1 \in \phi \end{aligned}\\ &\text { So, centre }(4,1)\\ &\begin{aligned} & \text { Radius }=1 \\ & \begin{aligned} \therefore \text { Area of circle } & =\pi r^2=\pi \times 1^2 \\ & =\pi \text { sq. units } \end{aligned} \end{aligned} \end{aligned} $

Line $x+2 y=1$ cuts the $X$-axis at $A$ and $Y$-axis at $B$.

Then, $A(1,0)$ and $B\left(0, \frac{1}{2}\right)$

A circle passes through $A$ and $B$ and origin.

So, equation of circle is

$ x^2+y^2+2 g x+2 f y=0 $

It passes through the point $A(1,0)$,

$ 1+0+2 g=0 \Rightarrow g=-\frac{1}{2} $

Also, passes through $\left(0, \frac{1}{2}\right)$

$ \begin{aligned} & & 0+\frac{1}{4}+0+f & =0 \\ \Rightarrow & & f & =-\frac{1}{4} \end{aligned} $

So, equation of circle is

$ x^2+y^2-x-\frac{y}{2}=0 $

So, tangent at $(0,0)$ is

$ \begin{array}{r} x x_1+y y_1-\left(\frac{x+x_1}{2}\right)-\frac{1}{2}\left(\frac{y+y_1}{2}\right)=0 \\ 0+0-\left(\frac{x+0}{2}\right)-\frac{1}{2}\left(\frac{y+0}{2}\right)=0 \end{array} $

⇒  $ 2 x+y=0 $

Perpendicular distance from $A(1,0)$ to tangent at

origin $=\left|\frac{2+0}{\sqrt{5}}\right|=\frac{2}{\sqrt{5}}$. Perpendicular distance from $B\left(0, \frac{1}{2}\right)$ to tangent at

$ \text { origin }=\left|\frac{0+\frac{1}{2}}{\sqrt{5}}\right|=\frac{1}{2 \sqrt{5}} $

Sum of perpendicular distances from $A$ and $B$ to tangent at origin

$ =\left(2+\frac{1}{2}\right) \frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{2} $

Centre $(-g,-f) \equiv\left(\frac{1}{2}, \frac{1}{4}\right)$

$ \begin{aligned} \text { Radius } & =\sqrt{g^2+f^2-c} \\ & =\sqrt{\frac{1}{4}+\frac{1}{16}-0}=\frac{\sqrt{5}}{4} \end{aligned} $

Diameter $=2 r=\frac{\sqrt{5}}{2}$

So, sum of perpendicular distances from $A$ and $B$ equal to the diameter of circle.

Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$

Equation of chord of contact of the point $P$ is : $x x_1+y y_1-a^2=0$

$ \begin{aligned} &\text { Qlies on chord of contact }\\ &\begin{aligned} & x_1 x_2+y_1 y_2-a^2=0 \\ & l_1=\text { Length of tangent drawn from } P \\ & =\sqrt{x_1^2+y_1^2-a^2} \\ & l_2=\text { Length of tangent drawn from } Q \\ & =\sqrt{x_2^2+y_2^2-a^2} \\ & \because P Q=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ & =\sqrt{x_1^2+y_1^2+x_2^2+y_2^2}-2\left(x_1 x_2+y_1 y_2\right) \\ & P Q=\sqrt{l_1^2+a^2+l_2^2+a^2-2 a^2} \\ & =\sqrt{l_1^2+l_2^2} \end{aligned} \end{aligned} $

$ \begin{aligned} & P A=3, Q B=2, A Q=3 \text { and } A B=5 \\ & \frac{P A}{A Q}=\frac{r_1}{r_2} \Rightarrow \frac{P A}{A Q}=\frac{3}{3}=1=\frac{r_1}{r_2} \\ \therefore \quad r_1 / r_2 & =1: 1 \end{aligned} $

$ \begin{aligned} &\text { Given circles }\\ &\begin{aligned} & S_1: x^2+y^2-6 x-4 y-23=0 \\ & C_1=(3,2), r_1=\sqrt{9+4+23}=6 \\ & S_2: x^2+y^2+2 x+2 y+1=0 \\ & C_2=(-1,-1), r_2=\sqrt{1+1-1}=1 \end{aligned} \end{aligned} $

$R$ divides $C_1$ and $C_2$ in ratio $6: 1$ externally.

$ \begin{aligned} R\left(\frac{6 \times(-1)-(3 \times 1)}{6-1},\right. & \left.\frac{6(-1)-2 \times 1}{6-1}\right) \\ & \equiv R\left(\frac{-9}{5}, \frac{-8}{5}\right) \end{aligned} $

So, equation of tangent through $R$ is

$ 5 y+8=5 m x+9 m $

$ \begin{aligned} & \Rightarrow \quad y+\frac{8}{5}=m\left(x+\frac{9}{5}\right) \\ & \Rightarrow 5 m x-5 y+9 m-8=0 \end{aligned} $

Distance from $C_2(-1,-1)$ to tangent $=1$

$ \begin{aligned} & \Rightarrow\left|\frac{-5 m+5+9 m-8}{\sqrt{25 m^2+25}}\right|=1 \\ & \Rightarrow \quad(4 m-3)^2=25 m^2+25 \\ & \Rightarrow \quad 16 m^2-24 m+9=25 m^2+25 \\ & \Rightarrow \quad 9 m^2+24 m+16=0 \Rightarrow(3 m+4)^2=0 \\ & \Rightarrow \quad m=\frac{-4}{3} \end{aligned} $

Now, $y+\frac{8}{5}=-\frac{4}{3}\left(x+\frac{9}{5}\right)$

$ \begin{aligned} \Rightarrow \quad 15 y+24 & =-20 x-36 \\ \Rightarrow \quad 20 x+15 y+60 & =0 \\ 4 x+3 y+12 & =0 \end{aligned} $

$ \begin{aligned} &\text { Given, } S_1 \text { : }\\ &\begin{array}{r} x^2+y^2-4 x-8 y+4=0 \\ C_1(2,4), r_1=\sqrt{4+16-4}=4 \\ S_2: x^2+y^2-8 x-12 y+16=0 \\ C_2(4,6), r_2=\sqrt{16+36-16}=6 \end{array} \end{aligned} $

$ \begin{aligned} &\text { Equation of common chord } S_2-S_1=0\\ &\begin{aligned} 4 x+4 y-12 & =0 \Rightarrow x+y=3 \\ C_1 M & =\left|\frac{2+4-3}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}} \end{aligned}\\ &\begin{aligned} P M & =\sqrt{\left(C_1 P\right)^2-\left(C_1 M\right)^2}=\sqrt{4^2-\left(\frac{3}{\sqrt{2}}\right)^2} \\ & =\sqrt{16-\frac{9}{2}}=\sqrt{\frac{23}{2}} \\ P Q & =2 P M=2 \times \sqrt{\frac{23}{2}}=\sqrt{46} \end{aligned} \end{aligned} $

Let the coordinate of point $P$ be $(h, k)$.

$ \begin{aligned} & (P A)^2=(h-1)^2+(k-1)^2 \\ & (P B)^2=(h+1)^2+(k-1)^2 \\ & (P C)^2=(h+1)^2+(k+1)^2 \\ & (P A)^2=(P B)^2+(P C)^2 \\ & \begin{array}{r} (h-1)^2+(k-1)^2=(h+1)^2+(k-1)^2 \\ +(h+1)^2+(k+1)^2 \end{array} \\ & \Rightarrow \quad h^2-2 h+1=h^2+2 h+1+h^2+1+2 h \\ & +k^2+2 k+1 \\ & \Rightarrow \quad h^2+k^2+6 h+2 k+2=0 \end{aligned} $

Locus of point $P$ is

$ x^2+y^2+6 x+2 y+2=0 $

Let equation of circle be

$ x^2+y^2+2 f x+2 g y+c=0 $

The circle passes through $(1,0)$

$ \begin{aligned} & 1+0+2 f+c=0 \\ & c=-2 f-1 \end{aligned} $

The equation of circle becomes

$ x^2+y^2+2 f x+2 g y-2 f-1=0 $

The circle passes through $(-1,1)$

$ \begin{gathered} 1+1-2 f+2 g-2 f-1=0 \\ 4 f-2 g=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

The circle also passes through $(2,-1)$

$ \begin{aligned} & 4+1+4 f-2 g-2 f-1=0 \\ & 2 f-2 g=-4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Subtracting Eq. (ii) from Eq. (i), we get

$ \begin{aligned} & & 2 f & =5 \\ \Rightarrow & & f & =\frac{5}{2} \end{aligned} $

Now, substitute $f=\frac{5}{2}$ in Eq. (i),

$ \begin{aligned} & 4 \times \frac{5}{2}-2 g=1 \\ \Rightarrow \quad & 10-2 g=1 \Rightarrow 2 g=9 \\ \Rightarrow \quad & g=\frac{9}{2} \\ c & =-2 f-1=-5-1=-6 \\ r & =\sqrt{g^2+f^2-c} \\ & =\sqrt{\frac{25}{4}+\frac{81}{4}+6} \\ & =\sqrt{\frac{25+81+24}{4}} \\ & =\sqrt{\frac{130}{4}}=\frac{\sqrt{130}}{2} \end{aligned} $

Given, equation of circle is

$ x^2+y^2-2 x-2 y+1=0 $

Centre of circle $=(1,1)$ and radius $=1$ unit

The point $B$ lies on the circle, the coordinate of $B$ is $(1+\cos \theta, 1+\sin \theta)$.

Given, coordinate of $A$ is $(0,-2)$

$ \begin{aligned} (A B)^2= & (1+\cos \theta)^2+(1+\sin \theta+2)^2 \\ = & 1+\cos ^2 \theta+2 \cos \theta+9+\sin ^2 \theta +6 \sin \theta \\ = & 11+2 \cos \theta+6 \sin \theta \end{aligned} $

As we know that

$ -\sqrt{a^2+b^2} \leq a \cos \theta+b \sin \theta \leq \sqrt{a^2+b^2} $

So,

$ \begin{aligned} &-\sqrt{4+36} \leq 2 \cos \theta+6 \sin \theta \leq \sqrt{4+36} \\ &-2 \sqrt{10}<2 \cos \theta+6 \sin \theta \leq 2 \sqrt{10} \end{aligned} $

Maximum value of $(A B)^2$ is $11+2 \sqrt{10}$.

Given, equation of circle is $x^2+y^2-10 x+14 y+46=0$ and equation of line is $3 x-5 y+6=0$ Let $P(\alpha, \beta)$ be the pole of line

$ 3 x-5 y+6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Equation of polar of point $P$ w.r.t. the circle, we get $x^2+y^2-10 x+14 y+46=0$ is

$ \begin{aligned} & \Rightarrow \alpha x+\beta y-5(x+\alpha)+7(y+\beta)+46=0 \\ & \Rightarrow x(\alpha-5)+y(\beta+7)-5 \alpha+7 \beta+46=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Comparing Eqs. (i) and (ii),

$ \begin{aligned} & & \frac{\alpha-5}{3}= & \frac{\beta+7}{-5} =\frac{-5 \alpha+7 \beta+46}{6}\\ \Rightarrow & & -5 \alpha+25 & =3 \beta+21 \\ \Rightarrow & & -5 \alpha & =3 \beta-4 \end{aligned} $

$ \begin{aligned} \therefore & \frac{\beta+7}{-5}=\frac{3 \beta-4+7 \beta+46}{6} \\ \Rightarrow & 6 \beta+42=-50 \beta+42 \times(-5) \\ \Rightarrow & 56 \beta=42 \times(-6) \\ \Rightarrow & \beta=\frac{42 \times(-6)}{56}=-\frac{9}{2} \\ \Rightarrow & -5 \alpha=3\left(-\frac{9}{2}\right)-4 \\ \Rightarrow & -5 \alpha=\frac{-27-8}{2} \\ \Rightarrow & -5 \alpha=-\frac{35}{2} \Rightarrow \alpha=\frac{7}{2} \\ \therefore & \alpha+\beta=\frac{7}{2}-\frac{9}{2}=-\frac{2}{2}=-1 \end{aligned} $

Given, $O(0,0)$ and $A(1,0)$ are the centres of 2 units circles $C_1$ and $C_3$.

The equation of circle passes through $Q(0,0)$ and $A(1,0)$ is

$ \begin{aligned} & (x-0)(x-1)+(y-0)(y-0) \\ & +\lambda\left|\begin{array}{ccc} x & y & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad x^2-x+y^2+\lambda y=0 \\ & \Rightarrow \quad x^2+y^2-x+\lambda y=0 \end{aligned} $

If it represents another unit circle $C_3$, its radius $=1$

$ \begin{aligned} & \text { Here, } g =-\frac{1}{2}, f=\frac{\lambda}{2}, c=0 \\ \Rightarrow & r =\sqrt{\frac{1}{4}+\frac{\lambda^2}{4}-0} \\ \Rightarrow & 1 =\frac{1}{4}+\frac{\lambda^2}{4} \Rightarrow \lambda^2=4-1 \\ \Rightarrow & \lambda = \pm \sqrt{3} \end{aligned} $

As the centre of $C_3$ lies above $X$-axis, $f$ is negative, so $\lambda$ is also negative.

$ \therefore \quad \lambda=-\sqrt{3} $

Equation of circle is

$ x^2+y^2-x-\sqrt{3} y=0 $

Since, $C_1$ and $C_3$ intersect and are of unit radius, their common tangent are parallel to the line joining the centres $(0,0)$ and $(1 / 2, \sqrt{3} / 2)$.

Let the equation of a common tangent be

$ \sqrt{3} x-y+k=0 $

It will touch $C_1$ if

$ \begin{aligned} \frac{|k|}{\sqrt{(\sqrt{3})^2+(-1)^2}} & =1 \\ k & = \pm 2 \end{aligned} $

∴ The required positive tangent make positive intercept on $Y$-axis and negative on $X$-axis and hence, the required equation is

$ \sqrt{3} x-y+2=0 $

Given, two equation of circles

$ x^2+y^2-16 x-20 y+164=r^2 $

and $x^2+y^2-8 x-14 y+29=0$

Here, $C_1(8,10)$,

$ r_1=\sqrt{64+100-164+r^2}=r $

and $C_2(4,7), r_2=\sqrt{16+49-29}=6$

If two circles with centre $C_1$ and $C_2$ and radius $r_1$ and $r_2$, respectively intersect at two points, then

$ \begin{aligned} & \left|r_1-r_2\right|

When $r>6$, then $r-6<5$ and $r+6>5 r<11$ and $r>1$

∴ The maximum integral value of $r=10$ When $r<6$, then

$ \begin{gathered} 6-r<5, r+6>5 \\ 6-511 \\ 1

$\therefore r$ is not possible in this case.

∴ The maximum integral value of $r$ is 10 .

Given, equation of circle is

$ x^2+y^2-6 x-12 y+1=0 $

Centre of circle is $(3,6)$ and radius of circle

$ \left(r_1\right)=\sqrt{9+36-1}=\sqrt{44} $

Let radius of another circle be $r$.

As the two circle cuts orthogonally

$ \begin{array}{rlrlrl} & & r_1^2+r_2^2 & =\left(C_1 C_2\right)^2 \\ & \Rightarrow & 44+r^2 & =(3+4)^2+(6-2)^2 \\ & \Rightarrow & 44+r^2 & =49+16 \\ & \Rightarrow & r^2 & =49+16-44=21 \\ \therefore & & r & =\sqrt{21} \end{array} $

Equation of incircle touches $x=0$ and $y=0$ is

$ (x-h)^2+(y-h)^2=h^2 $

Here, radius $=h$, and centre of circle is ( $h, h$ ).

The circle also touches the line

$ \begin{aligned} & \Rightarrow \quad 3 x+4 y-24=0 \\ & \Rightarrow \quad\left|\frac{3 h+4 h-24}{\sqrt{3^2+4^2}}\right|=h \end{aligned} $

$ \begin{gathered} 7 h-24= \pm 5 h \\ \Rightarrow \quad 7 h-24=5 h \text { and } 7 h-24=-5 h \\ h=12 \text { and } h=2 \\ (x-2)^2+(y-2)^2=2^2 \\ {[\because h=12 \text { is not acceptable }]} \end{gathered} $

$ \Rightarrow \quad x^2+y^2-4 x-4 y+4=0 $

Let $(2 \cos \theta, 2 \sin \theta)$ be the point on the circle $x^2+y^2=1$

$ \begin{array}{r} \therefore \text { Coordinate of } P=\left(2 \cos \frac{\pi}{4}, 2 \sin \frac{\pi}{4}\right) \\ =\left(2 \times \frac{1}{\sqrt{2}}, 2 \times \frac{1}{\sqrt{2}}\right)=(\sqrt{2}, \sqrt{2}) \end{array} $

Equation of tangent of circle is

$ \begin{aligned} & y=m x \pm r \sqrt{1+m^2} \\ & y=m x \pm \sqrt{1+m^2} \end{aligned} $

Point $(\sqrt{2}, \sqrt{2})$ lies on tangent

$ \begin{aligned} \sqrt{2} & =\sqrt{2} m+\sqrt{1+m^2} \\ \Rightarrow \quad \sqrt{2}(1-m) & = \pm \sqrt{1+m^2} \\ 2\left(1+m^2-2 m\right) & =1+m^2 \\ \Rightarrow m^2-4 m+1 & =0 \end{aligned} $

$ m=\frac{4 \pm \sqrt{16-4}}{2} \Rightarrow m=2 \pm \sqrt{3} $

Given, lines are

$ 5 x+6 y-34=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and  $ 2 x+y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned}Here,\,\, & l_1=5, m_1=6, n_1=34 \\ & l_2=2, m_2=1, n_2=c \end{aligned} $

And given circle

$ \begin{aligned} & x^2+y^2-8 x-10 y+25=0 \\ & \text { Here } \text {, centre }=(-g,-f)=(4,5) \\ & \text { and radius }=\sqrt{g^2+f^2-c} \\ & \qquad \begin{aligned} & =\sqrt{16+25-25}=4 \end{aligned} \end{aligned} $

Since, Eqs. (i) and (ii) are conjugate lines with respect to given circle

$ \begin{aligned} \therefore r^2\left(l_1 l_2+m_1 m_2\right)=\left(l_1 g+\right. & \left.m_1 f-n_1\right) \left(l_2 g+m_2 f-n_2\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow 16(10+6)=(-20-30+34)(-8-5-c) \\ & \Rightarrow \quad 16 \times 16=(-50+34)(-13-c) \end{aligned} $

$ \begin{array}{ll} \\ \Rightarrow & 16=13+c \\ \Rightarrow & c=3 \end{array} $

Now, line $2 x+y+c=2 x+y+3=0$

∴ Point $(1,-5)$ lines on this line.

Centre of circle $\left(O_1\right)=(-3,-4)$

Radius of circle

$ \left(r_1\right)=\sqrt{(-3)^2+(-4)^2-24}=1 $

Centre of circle $\left(O_2\right)=(3,4)$

Radius of circle $\left(r_2\right)=\sqrt{3^2+4^2-9}=4$

Centre of the similitude divides $O_1$ and $\mathrm{O}_2$ internally and externally in the ratio $r_1: r_2$

For Internally,

$ C_1=\left(\frac{3-12}{5}, \frac{4-16}{5}\right)=\left(\frac{-9}{5}, \frac{-12}{5}\right) $

For Externally,

$ \begin{aligned} C_2 & =\left(\frac{3+12}{-3}, \frac{4+16}{-3}\right)=\left(-5, \frac{-20}{3}\right) \\ C_1 C_2 & =\sqrt{\left(\frac{-9}{5}+5\right)^2+\left(\frac{-12}{5}+\frac{20}{3}\right)^2} \\ & =\sqrt{\frac{256}{25}+\frac{4096}{225}} \\ & =\sqrt{\frac{6400}{225}}=\frac{80}{15}=\frac{16}{3} \end{aligned} $

The radical axis of both circle is

$ \begin{gathered} S-S^{\prime}=0 \\ \Rightarrow x(2 g+6)+(2 f+4) y+(c-4)=0 \end{gathered} $

Compare it with given radical axis, we get $c=4$ and $\frac{1}{2 g+6}=\frac{1}{2 f+4}$

$ \begin{aligned} 2 g+6 & =2 f+4 \\ g & =f-1 \end{aligned} $

      As radius of first circle $=1$

$ \begin{aligned} \sqrt{g^2+f^2-c} & =1 \\ (f-1)^2+f^2-4 & =1 \\ \Rightarrow \quad 2 f^2+1-2 f-4 & =1 \\ f^2-f-2 & =0 \Rightarrow f=2,-1 \end{aligned} $

If $f=2$, then $g=1$ and $f+g=3$

If $f=-1$, then $g=-2$ are $f+g=-3$

Hence, $f+g= \pm 3$

The radical axis of circle

$ x^2+y^2-4 x-4 y+3=0 $

and $x^2+y^2+4 x-4 y+3=0$ is

$ -8 x=0 \Rightarrow x=0 $

The radius axis of circle

$ x^2+y^2+4 x-4 y+3=0 $

$ \begin{aligned}and\,\, & x^2+y^2+4 x+4 y+3=0 \text { is } \\ & \quad-8 y=0 \Rightarrow y=0 \end{aligned} $

∴ Centre of the required circle $=(0,0)$

Radius $=$ Length of the tangent from $(0,0)$ to each of the given circle.

$ \therefore \quad r=\sqrt{3} $

From the question, Equation of circle, $(x+2)^2+(y-3)^2=4$

Center of circle is $(-2,3)$

$\therefore B$ is the mid-point of $A Q[\because A Q=2 A B]$

$ \therefore B=\left(\frac{h}{2}, \frac{k+3}{2}\right) $

$\because B$ lies on the given circle

$ \begin{array}{lr} \therefore \left(\frac{h}{2}+2\right)^2+\left(\frac{k+3}{2}-3\right)^2=4 \\ \Rightarrow \frac{(h+4)^2}{4}+\frac{(k-3)^2}{4}=4 \\ \Rightarrow (h+4)^2+(k-3)^2=16 \\ \therefore \text { Locus: }(x+4)^2+(y-3)^2=16 \end{array} $

$ \begin{aligned} & x^2+y^2-6 x+4 y+12=0 \\ \Rightarrow \quad & (x-3)^2+(y+2)^2=1 \end{aligned} $

Equation of tangents

$ y+2=m(x-3) \pm 1 \sqrt{1+m^2} $

∵ It passes through $(1,-3)$.

$ \begin{aligned} & \therefore \quad-3+2=m(1-3) \pm \sqrt{1+m^2} \\ & \Rightarrow(2 m-1)= \pm \sqrt{1+m^2} \end{aligned} $

On squaring both sides,

$ \begin{aligned} & 4 m^2-4 m+1=1+m^2 \\ \Rightarrow \quad & 3 m^2-4 m=0 \Rightarrow m(3 m-4)=0 \\ \Rightarrow \quad & m=0, \frac{4}{3} \Rightarrow m_1=0, m_2=\frac{4}{3} \\ & 9\left(m_1^2+m_2^2\right)=9\left(0+\frac{16}{9}\right)=16 \end{aligned} $

$ \text { From the question } $

$ \begin{aligned} &\begin{gathered} x^2+y^2-8 x-10 y+5=0 \\ g=-4, f=-5 \text { and } c=5 \\ r=\sqrt{g^2+f^2-c}=\sqrt{36}=6 \end{gathered}\\ &\begin{aligned} & \therefore \text { Length of tangent } \\ & \quad=\sqrt{S_1}=\sqrt{4+9+16+30+5}=8 \\ & \therefore \quad \theta=2 \tan ^{-1}\left(\frac{r}{\sqrt{S_1}}\right)=2 \tan ^{-1}\left(\frac{6}{8}\right) \\ & \Rightarrow \quad \tan \frac{\theta}{2}=\frac{3}{4} \\ & \therefore \quad \tan \theta=\frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=\frac{2 \times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7} \end{aligned} \end{aligned} $

Let $c_1$ and $c_2$ are the centres and $r_1$ and $r_2$ are the radius of the two circles.

$ \begin{aligned} & \therefore c_1=(3,4), c_2=(-1,1) \Rightarrow r_1=4, r_2=1 \\ & \quad c_1 c_2=\sqrt{4^2+3^2}=5 \text { and } r_1+r_2=5 \\ & \Rightarrow c_1 c_2=r_1+r_2 \end{aligned} $

∴ Circles touch each other externally.

∴ Transverse common tangent is

$ \begin{aligned} \left(x^2+y^2-6 x-8 y+9\right) & -\left(x^2+y^2\right. +2 x-2 y+1)=0 \\ \Rightarrow-8 x-6 y+8=0 \Rightarrow & 4 x+3 y-4=0 \end{aligned} $

$ \begin{aligned} & \text { } c_1=(1,2), c_2=(4,6) \\ & \begin{aligned} r_1=3, r_2 & =3 \\ d=c_1 c_2 & =\sqrt{3^2+4^2}=5 \\ \therefore \cos \theta & =\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}=\frac{9+9-25}{2 \times 9} \\ & =-\frac{7}{18} \end{aligned} \end{aligned} $

$ \begin{aligned} \therefore|7 \sec \theta-18 \cos \theta| & =\left|7\left(\frac{-18}{7}\right)-18\left(\frac{-7}{18}\right)\right| \\ & =|-18+7|=11 \end{aligned} $

$ \begin{aligned} S & \equiv x^2+y^2+\alpha x+6 y=0 \\ S^{\prime} & \equiv x^2+y^2+2 \alpha x+\alpha y+6=0 \\ S^{\prime \prime} & \equiv x^2+y^2+6 \alpha x-\alpha y+3=0 \end{aligned} $

Radical axis are

$ \begin{array}{r} S-S^{\prime}=-\alpha x+(6-\alpha) y-6=0 \\ S-S^{\prime \prime}=-5 \alpha x+(6+\alpha) y-3=0 \\ S^{\prime}-S^{\prime \prime}=-4 \alpha x+2 \alpha y+3=0 \end{array} $

Radical centre is point of concurrence of three radical axis

$ \begin{aligned} &\begin{aligned} & \therefore \quad\left(0, \frac{3}{4}\right) \text { satisfies } S-S^{\prime}=0 \\ & \Rightarrow(6-\alpha) \frac{3}{4}-6=0 \Rightarrow 6-\alpha=8 \\ & \Rightarrow \alpha=-2 \\ & \therefore S^{\prime}=x^2+y^2-4 x-2 y+6=0 \\ & \text { Centre }=(2,1) \end{aligned}\\ &\text { ∴ Required distance }\\ &\begin{aligned} & =\sqrt{(2-0)^2+\left(1-\frac{3}{4}\right)^2} \\ & =\sqrt{\frac{65}{16}}=\frac{\sqrt{65}}{4} \end{aligned} \end{aligned} $

Slope of diameter, $A C=3 / 4$ Slope of $A O=3 / 4$

$ \frac{x_1+x_2}{2}=3 \Rightarrow x_1+x_2=6 $

$ \begin{gathered} \frac{y_1+y_2}{2}=2 \Rightarrow y_1+y_2=4 \\ \frac{y_2-y_1}{x_2-x_1}=\frac{3}{4} \Rightarrow\left(y_2-y_1\right)=\frac{3}{4}\left(x_2-x_1\right) \\ A C=50 \\ \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=50 \\ \quad\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2=2500 \\ \Rightarrow \quad \frac{9}{16}\left(x_2-x_1\right)^2+\left(x_2-x_1\right)^2=2500 \\ \Rightarrow \quad \frac{25}{16}\left(x_2-x_1\right)^2=2500 \\ \Rightarrow \quad\left(x_2-x_1\right)^2=1600 \\ \Rightarrow \quad\left(y_2-y_1\right)^2=\frac{9}{16}\left(x_2-x_1\right)^2 \\ =\frac{9}{16} \times 1600=900 \end{gathered} $

$ \begin{aligned} & \Rightarrow \quad\left(x_2-x_1\right)^2=1600 \\ & \Rightarrow \quad\left(x_1+x_2\right)^2-\left(4 x_1 x_2\right)=1600 \\ & \Rightarrow \quad 6^2-4 x_1 x_2=1600 \Rightarrow x_1 x_2=\frac{-1564}{4} \\ & \Rightarrow \quad\left(y_2-y_1\right)^2=900 \\ & \Rightarrow \quad\left(y_1+y_2\right)^2-4 y_1 y_2=900 \\ & \Rightarrow \quad 4^2-4 y_1 y_2=900 \Rightarrow y_1 y_2=-\frac{884}{4} \\ & \because \quad \frac{x_1 x_2}{y_1 y_2}=\frac{1564}{884}=\frac{23}{13} \end{aligned} $

Given circle passes through the points $(1,2),(3,4)$.

If its centre lines on the line

$ x-y+3=0 $

Then, slope of $A B=\frac{4-2}{3-1}=1$

Slope of line perpendicular to $A B=-1$

Mid-point of $A B=(2,3)$

Equation of line perpendicular to $A B$ and passing through mid-point of $A B$

$ \begin{aligned}& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y-3=-1(x-2) \\ &Line\,\,\,\,\, L_1: x+y=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, centre lies on line $x-y=-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Point of intersection of Eqs. (i) and (ii) is the centre of circle $x=1, y=4$

Centre, $O(1,4)$

Radius $=A O=\sqrt{0+2^2}=2$

$ \begin{aligned} & \text { Circle }: x^2+y^2-36=0 \\ & x^2+y^2=36 \\ & A P \cdot A Q=(A T)^2 \end{aligned} $

$ \begin{aligned} & A T=\text { Length of tangent }=\sqrt{x_1^2+y_1^2-36} \\ & =\sqrt{25+49-36}=\sqrt{38} \\ & A P \cdot A Q=(\sqrt{38})^2=38 \end{aligned} $