Circle

$ \begin{aligned} &\\ &\begin{array}{ll} \text { } & \frac{x}{3}+\frac{y}{4}=1 \\ \Rightarrow & 4 x+3 y=12 \\ \Rightarrow & 4 x+3 y-12=0 \end{array} \end{aligned} $

$ \begin{aligned} &\text { distance from the centre }(c, c) \text { to the line }\\ &\begin{aligned} & 4 x+3 y-12=0 \\ & d=\frac{|4 c+3 c-12|}{\sqrt{25}} \\ & =\frac{|4 c+3 c-12|}{5} \\ \Rightarrow & \frac{|7 c-12|}{5}=c \\ \Rightarrow & |7 c-12|=5 c \\ \Rightarrow & 7 c-12=5 c \text { and } 7 c-12=-5 c \\ c=6, c=1 & \end{aligned} \end{aligned} $

Circle I : $x^2+y^2-6 x-4 y+9=0$

Circle II : $x^2+y^2+2 x+2 y-7=0$

Point of contact : $(\alpha, \beta)$

$ \Rightarrow(x-3)^2+(y-2)^2=4 $

Center $=(3,2)$

$ (x+1)^2+(y+1)^2=9 $

centre $(-1,-1)$

Slope of line $(m)=\frac{3}{4}$

Equation of line $y-2=\frac{3}{4}(x-3)$

$ \Rightarrow 3 x-4 y-1=0 $

The point of contact ( $\alpha, \beta$ ) lies on the line $3 x-4 y-1=0$

$\Rightarrow$ Radii are $r_1=2$ and $r_2=3$

Ratio $=2: 3$

Using section formula

$ \begin{aligned} & \alpha=\frac{2(-1)+3(3)}{2+3}=\frac{7}{5}, \beta=\frac{2(-1)+3(2)}{2+3} \\ & \Rightarrow \beta=\frac{4}{5} \Rightarrow 7 \beta=7 \times \frac{4}{5}=4 \cdot \frac{7}{5} \\ & \therefore 7 \beta=4 \alpha \end{aligned} $

$x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0$

For the first circle, $g_1=-\lambda, f_1=-1$, $c_1=-7$ for the second circle $g_2=\frac{-4}{3}$,

$ f_2=\frac{29}{6}, c_2=0 $

Orthogonality condition.

$ \begin{aligned} & 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\ & \Rightarrow \quad 2(-\lambda)\left(\frac{-4}{3}\right)+2(-1)\left(\frac{29}{6}\right)=-7+0 \\ & \Rightarrow \quad \frac{8}{3} \lambda-\frac{29}{3}=-7 \\ & \Rightarrow \quad \frac{8}{3} \lambda=\frac{29}{3}-7 \\ & \Rightarrow \quad \frac{8}{3} \lambda=\frac{8}{3} \Rightarrow \lambda=1 \end{aligned} $

Here $Q$ is the inverse point of $P(-1,1)$ with respect to the circle

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

Centre of circle $O \equiv(1,-1)$

$ \begin{aligned} & \therefore \quad O P \cdot O Q=r^2 \\ & \Rightarrow O Q=\frac{r^2}{O P}=\frac{2}{\sqrt{(1+1)^2+(-1-1)^2}} \\ & =\frac{2}{\sqrt{8}}=\frac{1}{\sqrt{2}} \end{aligned} $

$O P, O Q$ are collinear

$\therefore$ Slope of $O P=-1$

$Q$ point lie on $O P Q$

$ \begin{array}{ll} \therefore & \frac{x-1}{\cos \theta}=\frac{y+1}{\sin \theta}=O Q \\ \Rightarrow & \frac{x-1}{\frac{-1}{\sqrt{2}}}=\frac{y+1}{\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}} \\ & x-1=\frac{-1}{2}, \quad y+1=\frac{1}{2} \\ & x=\frac{1}{2}, y=\frac{-1}{2} \\ \therefore & \quad Q \equiv\left(\frac{1}{2}, \frac{-1}{2}\right) \text { lie on line } \\ & \quad x-3 y-2=0 \end{array} $

Equation of circle passes through $(3,5),(5,5)$ and $(3,-3)$ is

Let equation be

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

This is passes through $(3,5),(5,5)$ and $(3,-3)$

$ \begin{gathered} 6 g+10 f+c=-34 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 10 g+10 f+c=50 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ 6 g-6 f+c=-18 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{gathered} $

From Eqs. (i), (ii) and (iii) $g=-4$,

$ f=-1 \text { and } c=0 $

Required equation of circle

$ x^2+y^2-8 x-2 y=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

and given circle is

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

are orthogonal

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

Given that length of common chord of two circles $=2 \sqrt{17}$, with same radii one circle is

$ \begin{aligned} \Rightarrow \quad & x^2+y^2+6 x+4 y-12=0 \\ & r=\sqrt{3^2+2^2+12}=5 \end{aligned} $

both circles have same radii $=5$

$ \begin{aligned} &\begin{aligned} & A B=2 \sqrt{17} \\ & A C=\sqrt{17} \end{aligned}\\ &\text { In } \triangle A O^{\prime} C\\ &\begin{aligned} O^{\prime} C & =O^{\prime} A^2-A C^2 \\ & =25-17 \Rightarrow O^{\prime} C=2 \sqrt{2} \\ O O^{\prime} & =4 \sqrt{2} \\ \Rightarrow \quad \cos \theta & =\frac{O^{\prime} A^2+O A^2-O^{\prime} O^2}{2\left(O A^{\prime}\right)(O A)} \\ \Rightarrow \quad \cos \theta & =\frac{25+25-32}{50} \Rightarrow \cos \theta=\frac{9}{25} \\ \theta & =\cos ^{-1}\left(\frac{9}{25}\right) \end{aligned} \end{aligned} $

Given that

$ S \equiv x^2+y^2-16=0 $

Let equation of

$ S^{\prime} \equiv(x-h)^2+(y-k)^2=5^2 $

Common chord $S_2-S_1=0$

$ 2 x h+2 k y+9=k^2+h^2 $

Given slope $=\frac{3}{4}$

$ \frac{-2 h}{2 k}=\frac{3}{4} \Rightarrow h=\frac{-3}{4} k $

The chord of maximum length is diameter

$\therefore(0,0)$ lies on common chord

$ \begin{aligned} & h^2+k^2=9 \\ & \frac{9}{16} k^2+k^2=9 \Rightarrow k= \pm \frac{12}{5} \\ & h=\mp \frac{9}{5} \\ & \mathrm{re} \equiv\left(\frac{-9}{5}, \frac{12}{5}\right),\left(\frac{9}{5}, \frac{-12}{5}\right) \end{aligned} $

Given equation of circles,

$ \begin{aligned} S & \equiv x^2+y^2+2 x-2 y+c=0 \\ S^{\prime} & =x^2+y^2-6 x-8 y+9=0 \\ \left(g_1, f_1\right) & \equiv-1,1, c_1=c \\ \left(g_2, f_2\right) & \equiv(3,4), c_2=9 \end{aligned} $

Angle between circle $S$ and $S^{\prime}$ be $\theta$

$ \begin{aligned} & \cos \theta=\frac{2\left(g_1 g_2+f_1 f_2\right)-\left(c_1+c_2\right)}{2 \sqrt{g_1^2+f_1^2-c_1} \sqrt{g_2^2+f_2^2-c_2}} \\ & \Rightarrow \quad \cos \theta=\frac{2(-3+4)-c-9}{2 \sqrt{2-c} \sqrt{25-9}} \\ & \Rightarrow \quad \frac{5}{16}=\frac{-c-7}{2 \sqrt{2-c}(4)} \\ & \Rightarrow \quad 5 \times \sqrt{2-c}=-2(c+7) \\ & \Rightarrow \quad 25(2-c)=4\left(c^2+14 c+49\right) \\ & \Rightarrow \quad 4 c^2+81 c+146=0 \\ & \Rightarrow \quad c=-2 \frac{-146}{8} \\ & \therefore \quad c=-2 \\ & \text { Radius }=\sqrt{1+1+2}=2 \end{aligned} $

Let centre $(h, k)$ and radius be $r$ of circle $S$

$ \because S \equiv(x-h)^2+(y-k)^2=r^2 $

Circle passes through origin

$ \because \quad r^2=h^2+k^2 $

Given, the line $x=2$ intersects the circle at two points let these points be ( $2, y_1$ ) and ( $2, y_2$ ) mid-points of these points is ( $2, k$ )

Now, the distance of $(2, k)$ to $\left(2, y_1\right)$ is 2 $y_1=k+2$ and $y_2=k-2$

( $2, k+2$ ) and ( $2, k-2$ ) are intersection point radius of the circle.

$ \begin{aligned} & h^2+k^2=(h-2)^2+(k-(k+2))^2 \\ & h^2+k^2=h^2+4-4 h+4 \\ & k^2=-4 h+8 \Rightarrow k^2+4 h=8 \end{aligned} $

Taking locus, we get

$ y^2+4 x=8 $

Given, line $2 x+y-10=0$ touches the circle at the point $(3,4)$.

$\because$ Perpendicular to this line passes through centre of circle.

Equation of line perpendicular to $2 x+y-10=0$ is

$ \begin{array}{ll} & x-2 y+\lambda=0 \\ \Rightarrow & 3-8+\lambda=0 \Rightarrow \lambda=5 \\ \therefore & x-2 y+5=0 \end{array} $

Centre of the circle be $C(2 k-5, k)$

Now radius $r^2=C P^2=C Q^2$

$ r^2=(2 k-8)^2+(k-4)^2=(2 k-6)^2+(k+2)^2 $

On solving this, we get $k=2$

$\because$ Centre $(-1,2)$ and radius $=\sqrt{20}$

Equation of circle be

$ (x+1)^2+(y-2)^2=20 $

Clearly ( $-5,4$ ) lies on the circle.

$(a, b)$ is the common point for the two given circle

$\eqalign{ & {a^2} + {b^2} - 4a + 4b - 1 = 0\,\,\,\,\,\,\,\,\,...(i) \cr & \,\,\,\,{a^2} + {b^2} + 2a - 4b + 1 = 0\,\,\,\,\,\,\,\,...(ii) \cr & \underline { - \,\,\,\,\,\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\, + \,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr & Subtrating\,Eqs.(ii)form(i),we\,get \cr} $

$ \begin{aligned} & -6 a+8 b-2=0 \\ & -3 a+4 b-1=0 \\ & 4 b=3 a+1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (i) and (iii), we get

$ \begin{aligned} & a^2+\left(\frac{3 a+1}{4}\right)^2-4 a+3 a+1-1=0 \\ & \Rightarrow 25 a^2-10 a+1=0 \\ & \quad(5 a-1)^2=0 \\ & \quad a=\frac{1}{5} \text { and } b=\frac{2}{5} \end{aligned} $

Now, $a^2+b^2=\frac{1}{25}+\frac{4}{25}=\frac{5}{25}=\frac{1}{5}$

$ \begin{aligned} &\text { We have the circle }\\ &\begin{gathered} x^2+y^2+4 x+4 y+c=0 \\ C=\text { Centre }=(-2,-2) \\ A C=\text { radius }=\sqrt{8-C} \end{gathered} \end{aligned} $

$ \begin{aligned} &\text { Angle between tangents }\\ &\begin{aligned} & \cos 2 \theta=\frac{7}{16}, C P=\sqrt{16+16}=4 \sqrt{2} \\ & \because \sin \theta=\frac{3}{4 \sqrt{2}} \\ & \text { Using } \sin \theta=\frac{r}{C P}=\frac{\sqrt{8-C}}{4 \sqrt{2}}=\frac{3}{4 \sqrt{2}} \\ & \because \quad 8-C=9 \\ & \quad C=-1 \end{aligned} \end{aligned} $

Now, another circle is possible when angles be supplementary

$\because \theta$ is changed by $\pi-2 \theta$

$ \begin{aligned} & \because \quad \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta \\ & \because \quad \cos \theta=\sqrt{\frac{1+\cos 2 \theta}{2}}=\frac{\sqrt{23}}{4 \sqrt{2}} \\ & \because \quad \cos \theta=\frac{\sqrt{8-C}}{4 \sqrt{2}}=\frac{\sqrt{23}}{4 \sqrt{2}} \\ & \because \quad 8-c=23 \\ & -c=-15 \end{aligned} $

$\because$ Now, sum of all possible values of $C$.

$ =-1-15=-16 $

We have the circle $S$

$ S=x^2+y^2+2 g x+4 y+1=0 $

and the second circle be $S_1$,

$ S_1=x^2+y^2-2 x-3=0 $

$\because$ Centre $(1,0)$ and radius $=2$

Now, common chord of circles $S$ and $S_1$ is

$ \begin{aligned} & S-S_1=0 \\ & (2 g+2) x+4 y+4=0 \end{aligned} $

If a circle bisects the circumference of another circle, then common chord is a diameter of the second circle.

$ \begin{array}{ll} & 2 g+2+4=0 \Rightarrow g=-3 \\ \because \quad & S=x^2+y^2-6 x+4 y+1=0 \end{array} $

Centre $(3,-2)$

Radius $=\sqrt{9+4-1}=\sqrt{12}$

Circle 1 given $x^2+y^2=4$

Circle 2 is given,

$ x^2+y^2-6 x-6 y+14=0 $

Centre $(3,3)$ radius $=2$

Let $P\left(x_1, y_1\right)$ be a point on circle 1 .

$\because$ Equation of chord of contact $A B$ is

$ \begin{aligned} & x x_1+y y_1-3\left(x+x_1\right)-3\left(y+y_1\right)+14=0 \\ & \left(x_1-3\right) x+\left(y_1-3\right) y-3 x_1-3 y_1+14=0 \end{aligned} $

Now, equation of circle passing through $P, A$ and $B$ is

$ \begin{aligned} x^2+y^2-6 x-6 y+14 +\lambda\left[\left(x_1-3\right) x+\left(y_1-3\right) y-3 x_1\right. & \left.-3 y_1+14\right]=0 \end{aligned} $

We have this circle passes through $P\left(x_1, y_1\right)$

$ \begin{array}{r} \\\Rightarrow x_1^2+y_1^2-6 x_1-6 y_1+14+\lambda\left[\left(x_1-3\right) x_1\right. \\ \left.+\left(y_1-3\right) y_1-3 x_1-3 y_1+14\right]=0 \\ \Rightarrow x_1^2+y_1^2-6 x_1-6 y_1+14+\lambda\left[x_1^2+y_1^2\right. \\ \left.-3 x_1-3 y_1-3 x_1-3 y_1+14\right]=x \end{array} $

Put $x_1^2+y_1^2=4$

$ \begin{gathered} \Rightarrow 4-6 x_1-6 y_1+14 \\ +\lambda\left(4-6 x_1-6 y_1+14\right)=0 \\ \Rightarrow 18-6 x_1-6 y_1+14+\lambda\left(18-6 x_1\right. \\ \left.-6 y_1\right)=0 \end{gathered} $

$ \begin{aligned} & \because 1+\lambda=0 \\ & \lambda=-1 \end{aligned} $

$\Rightarrow$ Equation of circle be

$ \begin{aligned} & \left(x^2+y^2-6 x-6 y+14\right) \\ & -\left[\left(x_1-3\right) x+\left(y_1-3\right) y\right. \\ & \left.-3 x_1-3 y_1+14\right]=0 \\ & \Rightarrow x^2+y^2-\left(3+x_1\right) x-\left(3+y_1\right) y \\ & +3 x_1+3 y_1=0 \end{aligned} $

Centre $\left(\frac{3+x_1}{2}, \frac{3+y_1}{2}\right)$

$ h=\frac{3+x_1}{2} \text { and } k=\frac{3+y_1}{2} $

$ \begin{aligned} & x_1=2 h-3 \quad y_1=2 k-3 \\ & x_1^2+y_1^2=4 \\ & (2 h-3)^2+(2 k-3)^2=4 \\ & 4 h^2+4 k^2-12 h-12 k+14=0 \\ & h^2+k^2-3 h-3 k+\frac{7}{2}=0 \end{aligned} $

$\because$ Taking locus of $(h, k)$, we get

$ 2 x^2+2 y^2-6 x-6 y+7=0 $

Let the ends of diameter of the given circle $x^2+y^2=4$ is

$(2 \cos \theta, 2 \sin \theta)$, then other must be ( $-2 \cos \theta,-2 \sin \theta$ )

Then, length of perpendicular from the extremities of diameter to the line $x+y+1=0$ is

$ \begin{aligned} & \frac{2(\cos \theta+\sin \theta)+1}{\sqrt{2}} \\ & \times \frac{-2(\cos \theta+\sin \theta)+1}{\sqrt{2}}=P(\text { say }) \\ & \Rightarrow P=\frac{1-4(\cos \theta+\sin \theta)^2}{2} \\ & =\frac{1-4(1+\sin 2 \theta)}{2} \\ & \Rightarrow P=\frac{-3-4 \sin 2 \theta}{2} \\ & \Rightarrow \quad \frac{d P}{d \theta}=\frac{1}{2}(-4) \cos 2 \theta \cdot 2=0 \\ & \Rightarrow \quad \cos 2 \theta=0 \Rightarrow 2 \theta=\frac{\pi}{2} \\ & \quad \theta=\frac{\pi}{4} \end{aligned} $

Hence, the points are $(\sqrt{2}, \sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$

Let the centre of circle be $(h, k)$ and radius be $r$.

$ \begin{aligned}Clearly,\, P M & =k \\ P N & =h \\ A M & =4, C N=3 \\ r^2 & =4^2+k^2 \\ r^2 & =3^2+h^2 \end{aligned} $

$ \begin{aligned} \because & 16+k^2=9+h^2 \\ & h^2-k^2-7=0 \end{aligned} $

Taking locus of $P(h, k)$, we get

$ x^2-y^2-7=0 $

Equation of tangent to the circle,

$ x^2+y^2=9 \text { is } y=m x+3 \sqrt{1+m^2} $

Tangent passes through the point $(3,4)$,

$ \begin{aligned} & (4-3 m)^2=9\left(1+m^2\right) \\ & \Rightarrow \quad 16+9 m^2-24 m=9+9 m^2 \\ & \Rightarrow \quad 24 m=7 \\ & \quad m=\frac{7}{24} \end{aligned} $

$ \text { } \begin{aligned} S & =x^2+y^2+2 k x+4 y-3=0 \\ C_1 & =(-k,-2) \text { radius } r_1=\sqrt{k^2+4+3} \\ S^{\prime} & =x^2+y^2-4 x+2 k y+9=0 \\ C_2 & =(2,-k) \text { radius }=\sqrt{4+k^2-9} \\ & =\sqrt{k^2-5} \end{aligned} $

Now, angle between two circles $S=0$ and $S^{\prime}=0$ is

$ \cos \left(180^{\circ}-\theta\right)=\frac{\left(r_1^2+r_2^2-\left(d^2\right)\right.}{2 r_1 r_2} $

where $r_1, r_2$ are the radii and $d$ is the distance between their centres.

$ \begin{aligned} & =\frac{k^2+7+k^2-5-(k+2)^2-\left(k-2)^2\right.}{2 \sqrt{k^2+7} \sqrt{k^2-5}} \\ & \Rightarrow-\cos \theta=\frac{2 k^2+2-\left(2 k^2+8\right)}{2 \sqrt{k^2+7} \sqrt{k^2-5}} \\ & =\frac{-6}{2 \sqrt{k^2+7} \sqrt{k^2-5}} \\ & \Rightarrow \quad \cos \theta=\frac{3}{\sqrt{k^2+7} \sqrt{k^2-5}}=\frac{3}{8} \\ & \Rightarrow \quad\left(k^2+7\right)\left(k^2-9\right)=64 \\ & \Rightarrow \quad k^4+2 k^2-99=0 \\ & \Rightarrow \quad\left(k^2+11\right)\left(k^2-9\right)=0 \Rightarrow k^2=9 \\ & \Rightarrow \quad k= \pm 3 \end{aligned} $

Centre of $S^{\prime}=0$ lies in I quadrant

$ [\therefore k=-3] $

Now, radical axis of $S=0$ and $S^{\prime}=0$ is

$ \Rightarrow \quad \begin{aligned} & x(2 k+4)+y(4-2 k)-12=0 \\ & -2 x+10 y-12=0 \\ & x-5 y+6=0 \end{aligned} $

Let $(h, k)$ be the mid-point of a chord of the circle $x^2+y^2=25$.

Then, equation of the chord is $h x+k y-25$

$ =h^2+k^2-25 \Rightarrow h x+k y=h^2+k^2 $

If this is parallel to $x-2 y+3=0$, then $-\frac{h}{k}=\frac{1}{2} \Rightarrow 2 h+k=0$

So, the locus of $(h, k)$ is $2 x+y=0$.

The point $(2 a, a+1)$ will be an interior point of the larger segment of the circle.

$ x^2+y^2-2 x-2 y-8=0 $

if the point $(2 a, a+1)$ is an interior point and the point $(2 a, a+1)$ and the centre $(1,1)$ are on the same side of the chord $x-y+1=0$.

$ \begin{aligned} & \therefore(2 a)^2+(a+1)^2-2(2 a)-2(a+1)-8 < 0 \\ & \text { and }(2 a-a-1+1)(1-1+1) > 0 \\ & \Rightarrow \quad 5 a^2-4 a-9 < 0 \text { and } a > 0 \\ & \Rightarrow \quad(5 a-9)(a+1)<0 \text { and } a > 0 \\ & \Rightarrow \quad-1 < a <\frac{9}{5} \text { and } a > 0 \\ & \Rightarrow \quad a \in\left(0, \frac{9}{5}\right) \end{aligned} $

$\begin{aligned} & 3 x+5 y=1 \\ & (2+c) x+5 c^2 y=1 \\ & 3 c^2 x+5 c^2 y=c^2 \end{aligned}$

Subtracting

$\begin{aligned} & \left(2+c-3 c^2\right) x=1-c^2 \\ & x_c=\frac{1-c^2}{2+c-3 c^2}=\frac{(1-c)(1+c)}{(1-c)(3 c+2)}=\frac{c+1}{3 c+2} \\ & y=\frac{1-3 x}{5}=\frac{1-3\left(\frac{c+1}{3 c+2}\right)}{5} \\ & =\frac{3 c+2-3 c-3}{5(3 c+2)} \\ & y_c=\frac{-1}{5(3 c+2)} \\ & \lim _{c \rightarrow 1} x_c=\frac{2}{5}=h \\ & \lim _{c \rightarrow 1} y_c=\frac{-1}{25}=k \end{aligned}$

Equation of circle is

$25 x^2+25 y^2-20 x+2 y-60=0$

$\begin{aligned} & \frac{x+4}{1}=\frac{y-5}{2}=\frac{-2(4)}{5} \\ & \Rightarrow \quad x=-4-\frac{8}{5}=-\frac{28}{5}, y=5-\frac{16}{5}=\frac{9}{5} \\ & \therefore \quad \text { Image is }\left(\frac{-28}{5}, \frac{9}{5}\right) \end{aligned}$

Image lies on circle $(x+4)^2+(y-3)^2=r^2$

$\begin{aligned} & \left(\frac{-28}{5}+4\right)^2+\left(\frac{9}{5}-3\right)^2=r^2 \\ & \Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \\ & \Rightarrow r=2 \end{aligned}$

$\begin{aligned} & C_1 \rightarrow(x-\alpha)^2+(y-\beta)^2=r_1^2 \\ & C_2 \rightarrow(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2 \end{aligned}$

Point $P$ divide the line segment internally $C_1 C_2$ in the ratio 2 : 1

$\begin{aligned} & \frac{\alpha \times 1+8 \times 2}{1+2}=6, \alpha=2 \\ & \frac{\beta \times 1+\frac{15}{\alpha} \times 2}{1+2}=6, \beta=3 \\ & r_1=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{25}=5 \\ & r_2=\sqrt{(8-6)^2+\left(\frac{15}{\alpha}-6\right)^2}=\frac{5}{2} \\ & \alpha+\beta+4\left(r_1^2+r_2^2\right)=2+3+4\left(5^2+\left(\frac{5}{2}\right)^2\right) \\ & =5+4\left(\frac{125}{4}\right) \\ & =130 \\ \end{aligned}$

Slope of $B F=\left(m_{B F}\right)=\frac{3-1}{8-6}=1$

$\begin{aligned} & m_{A C}=\frac{k+2}{h-5}=-1(A C \perp B F) \\ & \Rightarrow k+2=5-h \\ & \Rightarrow k=3-h \quad \ldots \text { (i) } \\ & m_{A D}=\frac{1+2}{6-5}=3 \\ & m_{B C}=\frac{k-3}{h-8}=-\frac{1}{3}(A D \perp B C) \\ & \Rightarrow 3 k-9=8-h \quad \ldots \text { (ii) } \end{aligned}$

From (i) & (ii)

$h=-4, k=7$

Now, $h^2+k^2=16+49=65$

$h^2+k^2-65=0$

Locus is $x^2+y^2-65=0$

Inradius of $\triangle A B C=r=\frac{\Delta}{s}=\frac{\frac{\sqrt{3}}{4} \times(12)^2}{18}$

$r=2 \sqrt{3}$

Side length of square is $a$, then $a^2=2 r^2$

$\Rightarrow a^2=24$

Area of square, $m=24$

Perimeter of square, $n=4 \sqrt{24}$

$\begin{aligned} & \Rightarrow m+n^2=24+384 \\ & =408 \end{aligned}$

$\begin{gathered} C_1: x^2+y^2-2(x+y)+1=0 \\ C_2:(x+1)^2+y^2=(2)^2 \\ x^2+y^2+2 x-3=0 \end{gathered}$

Common chord is

$\begin{aligned} & C_1-C_2=0 \\ & \Rightarrow 2 x+y-2=0 \end{aligned}$

also, this line intersects the $y$-axis at the point

$\begin{aligned} & P(y, 0) . \\ & \Rightarrow y=2 \end{aligned}$

$P(2,0)$

Distance of point $P$ from $(1,1)$ is

$\begin{aligned} & d=\sqrt{(2-1)^2+(0-1)^2} \\ & =\sqrt{1^2+1^2} \\ & d=\sqrt{2} \\ & \Rightarrow d^2=2 \\ \end{aligned}$

$C F=4 \sqrt{2}-2 \sqrt{2}=2 \sqrt{2}=r+r \sqrt{2}$

$\begin{aligned} & \Rightarrow \quad(2-r) \sqrt{2}=r \\ & \Rightarrow \quad \sqrt{2}=\left(\frac{r}{2-r}\right) \Rightarrow 2=\frac{r^2}{(2-r)^2} \\ & \Rightarrow 2\left(r^2-4 r+4\right)=r^2 \\ & \Rightarrow r^2-8 r+8=0 \end{aligned}$

Shortest distance of circle $C$ form $(5,5)$

$\begin{aligned} & =\sqrt{9+16}-1 \\ & =5-1=4 \end{aligned}$

$\text { Let the line by } x+y=\lambda$

$\begin{aligned} & \therefore\left|\frac{\lambda}{\sqrt{2}}\right|=3 \\ & \therefore \quad \lambda= \pm 3 \sqrt{2} \end{aligned}$

$\therefore$ distance between lines $x+y=2$ and $x+y=3 \sqrt{2}$ is

$\frac{3 \sqrt{2}-2}{\sqrt{2}}=3-\sqrt{2}$

One side of square is $y=x+k$ Distance of $(5,3)$ to the line $y=x+k$ is

$\begin{aligned} & \frac{|3-5-k|}{\sqrt{2}}=\sqrt{2} \\ & =|-2-k|=2 \\ & \Rightarrow k=0 \text { or } k=-4 \end{aligned}$

So lines are $y=x$ and $y=x-4$

Now, solving these lines with circle

$\begin{aligned} y= & x \text { and } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & 2 x^2-16 x+30=0 \\ \Rightarrow & x=3, y=3 \\ & x=5, y=5 \\ & y=x-4 \text { and } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & x=5, y=1 \\ & x=7, y=3 \\ & \sum_{i=1}^4 x_i^2+y_i^2=9+9+25+25+25+1+49+9 \\ = & 152 \end{aligned}$

$\begin{aligned} & (y-2)=m(x-8) \\ & \Rightarrow x \text {-intercept } \\ & \Rightarrow\left(\frac{-2}{m}+8\right) \\ & \Rightarrow y \text {-intercept } \\ & \Rightarrow(-8 \mathrm{~m}+2) \\ & \Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{~m}}+8-8 \mathrm{~m}+2 \\ & \mathrm{f}^{\prime}(\mathrm{m})=\frac{2}{\mathrm{~m}^2}-8=0 \\ & \Rightarrow \mathrm{m}^2=\frac{1}{4} \\ & \Rightarrow \mathrm{m}=\frac{-1}{2} \\ & \Rightarrow \mathrm{f}\left(\frac{-1}{2}\right)=18 \\ & \Rightarrow \text { Minimum }=18 \end{aligned}$

$\begin{aligned} & 2 x+3 y=12 \\ & 3 x-2 y=5 \end{aligned}$

$\begin{aligned} & 13 x=39 \\ & x=3, y=2 \end{aligned}$

Center of given circle is $(5,-2)$

Radius $\sqrt{25+4-13}=4$

$\begin{aligned} & \therefore \mathrm{CM}=\sqrt{4+16}=5 \sqrt{2} \\ & \therefore \mathrm{CP}=\sqrt{16+20}=6 \end{aligned}$

If two circles intersect at two distinct points

$\begin{aligned} & \Rightarrow\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & |\mathrm{r}-2|<\sqrt{9+16}<\mathrm{r}+2 \\ & |\mathrm{r}-2|<5 \text { and } \mathrm{r}+2>5 \\ & -5<\mathrm{r}-2<5 \quad \mathrm{r}>3 ~\text{......... 2} \end{aligned}$

$-3<\mathrm{r}<7\quad$ .... (1)

From (1) and (2)

$3<\text { r }<7$

$(2 k, 3 k)$ will lie on circle whose diameter is $A B$.

$\begin{aligned} & (x-1)(x)+(y-1)(y)=0 \\ & x^2+y^2-x-y=0 \quad \text{.... (i)} \end{aligned}$

Satisfy (2k, 3k) in (i)

$\begin{aligned} & (2 k)^2+(3 k)^2-2 k-3 k=0 \\ & 13 k^2-5 k=0 \\ & k=0, k=\frac{5}{13} \\ & \text { hence } k=\frac{5}{13} \end{aligned}$

Consider centre as $\mathrm{P}(0, \alpha), \alpha>0$

$\left|\frac{2(0)-\alpha}{\sqrt{5}}\right|=\mathrm{r}$

$\begin{aligned} & |-\alpha|=\sqrt{5} r \\ & \alpha=\sqrt{5} r \\ & \therefore \alpha+r=5+\sqrt{5} \\ & \sqrt{5} r+r=\sqrt{5}(\sqrt{5}+1) \\ & r=\sqrt{5}, \alpha=5 \\ & \therefore P(0,5) \end{aligned}$

Foot of perpendicular from $P$ to line $2 x-y=0$

$\begin{aligned} & \frac{x-0}{2}=\frac{y-5}{-1}=\frac{-(2(0)-5)}{5}=1 \\ & x=2, y=4 \quad A_1(2,4) \end{aligned}$

Let $\mathrm{B}(\mathrm{p}, \mathrm{q}) \quad \therefore \frac{\mathrm{p}+2}{2}=0, \frac{\mathrm{q}+4}{2}=5$

$\therefore \mathrm{p}=-2, \mathrm{q}=6 \quad B(-2,6)$

$ P\left(\frac{-5+6}{3}, \frac{3-14}{3}\right)=\left(\frac{1}{3}, \frac{-11}{3}\right) $

$ Q\left(\frac{-10+3}{3}, \frac{6-7}{3}\right)=\left(-\frac{7}{3},-\frac{1}{3}\right) $

Slope of $ P R=\frac{K+\frac{11}{3}}{h-\frac{1}{3}}=\frac{3 K+11}{3 h-1} $

Slope of $ Q R=\frac{K+\frac{1}{3}}{h+\frac{7}{3}}=\frac{3 K+1}{3 h+7} $

$ P R \perp Q R $

$ \frac{3 K+11}{3 h-1} \times \frac{3 K+1}{3 h+7}=-1 $

$ \begin{array}{l} (3 K+11)(3 K+1)=-(3 h-1)(3 h+7) \\\\ 9 K^{2}+36 K+11=-\left[9 h^{2}+18 h-7\right] \\\\ 9 h^{2}+18 h-7+9 K^{2}+36 K+11=0 \\\\ 9 h^{2}+9 K^{2}+18 h+36 K+4=0 \\\\ h^{2}+K^{2}+2 h+4 K+\frac{4}{9}=0 \end{array} $

Taking locus, we get

$ \begin{array}{l} \Rightarrow x^{2}+y^{2}+2 x+4 y+\frac{4}{9}=0 \\\\ \therefore \text { radius }=\sqrt{1+4-\frac{4}{9}}=\sqrt{\frac{41}{9}}=\frac{\sqrt{41}}{3} \end{array} $

$A B=\sqrt{(1)^1+(l)^2}=\sqrt{2}$ units

$ \because D $ is the mid point of $ A B $

So, $ A D=1 / \sqrt{2} $ units

$ \therefore \quad D \equiv\left(\frac{3}{2}, \frac{3}{2}\right) $

Now, shortest distance $ =\sqrt{\frac{9}{4}+\frac{9}{4}} $

$ =\frac{3 \sqrt{2}}{2} \text { units } $

$ \begin{aligned} & \tan \theta = \frac{A D}{S D} \Rightarrow \tan \theta=\frac{1}{\sqrt{2}} \times \frac{2}{3 \sqrt{2}}=\frac{1}{3} \\\\ \therefore \quad & \theta=\angle A C B=\angle A S D=\tan ^{-1}\left(\frac{1}{3}\right) \end{aligned} $

Hence, the angle substances by $ A B $ at the third vertex is $ \tan ^{-1}\left(\frac{1}{3}\right) $.

To find a possible value of $ g $ for a circle described by the equation $ x^{2} + y^{2} + 2gx + 2fy + c = 0 $, which passes through the points $(1,1)$ and $(2,0)$ and touches the line $3x - y - 1 = 0$, we begin with the following steps:

Pass through Points:

For point $(1,1)$:

$ 1^2 + 1^2 + 2g \cdot 1 + 2f \cdot 1 + c = 0 \\ 2 + 2g + 2f + c = 0 \quad \ldots \text{(i)} $

For point $(2,0)$:

$ 2^2 + 0^2 + 2g \cdot 2 + 2f \cdot 0 + c = 0 \\ 4 + 4g + c = 0 \quad \ldots \text{(ii)} $

Solve the System of Equations:

Subtract equation (ii) from (i):

$ (2 + 2g + 2f + c) - (4 + 4g + c) = 0 \\ -2 - 2g + 2f = 0 \\ \Rightarrow 2f = 2g + 2 \\ f = g + 1 $

From equation (ii), solve for $ c $:

$ 4 + 4g + c = 0 \\ c = -4g - 4 $

Condition for Tangency:

Given the line $3x - y - 1 = 0$ touches the circle, use the tangency condition:

$ \frac{-3g + f - 1}{\sqrt{3^2 + (-1)^2}} = \sqrt{g^{2} + f^{2} - c} $

$ \frac{-3g + f - 1}{\sqrt{10}} = \sqrt{g^{2} + f^{2} - c} $

Substitute $ f = g + 1 $ and $ c = -4g - 4 $:

$ (-3g + (g + 1) - 1)^2 = 10(g^2 + (g + 1)^2 - (-4g - 4)) $

Simplifying:

$ (-3g + g)^2 = 10((g + 1)^2 - (-4g - 4)) \\ 4g^2 = 10(g^2 + 2g + 1 + 4g + 4) \\ 4g^2 = 10(2g^2 + 6g + 5) \\ 4g^2 = 20g^2 + 60g + 50 $

Simplifying further:

$ 16g^2 + 60g + 50 = 0 \\ 8g^2 + 30g + 25 = 0 $

Solving this quadratic equation gives:

$ g = -\frac{5}{2} $

Let the equation of the circle be:

$ x^{2} + y^{2} + 2gx + 2fy + c = 0 $

The circle passes through the points $ (2,0) $ and $ (1,2) $.

From the point $(2,0)$:

$ 4 + 4g + c = 0 \quad \ldots \text{(i)} $

From the point $(1,2)$:

$ 5 + 2g + 4f + c = 0 \quad \ldots \text{(ii)} $

Subtracting equation (i) from equation (ii):

$ (-1) + 2g - 4f = 0 $

$ 2g - 4f = 1 \quad \ldots \text{(iii)} $

Denoting the power of the point $(0,2)$ with respect to this circle as 4:

$ 0 + 4 + 4f + c = 4 $

$ 4f + c = 0 $

$ c = -4f \quad \ldots \text{(iv)} $

From equation (iii):

$ 2g - 4f = 1 $

$ 4g - 4f = -4 $

Solving for $f$:

$ -2f = -3 \Rightarrow f = -\frac{3}{2} $

Substituting the value of $f$ into equation (iv):

$ c = -4\left(-\frac{3}{2}\right) = 6 $

Solving for $g$:

$ g = f - 1 = -\frac{3}{2} - 1 = -\frac{5}{2} $

Now, calculate the radius:

$ \text{Radius} = \sqrt{g^2 + f^2 - c} = \sqrt{\frac{25}{4} + \frac{9}{4} - 6} $

$ = \sqrt{\frac{17}{2} - 6} = \sqrt{\frac{5}{2}} $

The given circle equation is:

$ x^2 + y^2 + 2gx + 2fy - 8 = 0 $

The center of this circle is at $(-g, -f)$.

We are told that the line $x - 2y - 6 = 0$ is a normal to the circle, which means it passes through the center ($-g, -f$) of the circle. Therefore, we have:

$ -g + 2f = 6 \quad \ldots \text{(i)} $

Additionally, the line $y = 2$ is tangent to the circle. The general equation for the distance from a point to a line is given by:

$ \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} $

The distance from the center $(-g, -f)$ to the line $y = 2$ is:

$ \frac{|f - 2|}{1} = \sqrt{g^2 + f^2 + 8} $

Squaring both sides yields:

$ (f - 2)^2 = g^2 + f^2 + 8 $

Simplifying further:

$ f^2 - 4f + 4 = g^2 + f^2 + 8 $

$ -4f + 4 = g^2 + 8 $

Rearranging, we get:

$ 4f = g^2 + 4 \quad \ldots \text{(ii)} $

Substitute $g = 4$ and $g = -2$ into the equation obtained above:

For $g = 4$,

$ 4f = 16 + 4 = 20 $

$ f = 5 $

For $g = -2$,

$ 4f = 4 + 4 = 8 $

$ f = 2 $

Calculating the radius:

For $f = 5$,

$ \text{Radius} = \sqrt{16 + 25 + 8} = \sqrt{49} = 7 $

For $f = 2$,

$ \text{Radius} = \sqrt{4 + 4 + 8} = \sqrt{16} = 4 $

Thus, the radius of the circle can be $4$.

Given:

$ y = -x - 1 $

Substitute $ y = -x - 1 $ into the circle's equation:

$ x^2 + (-x - 1)^2 - 4x + 2(-x - 1) - 4 = 0 $

Simplify:

$ x^2 + (x^2 + 2x + 1) - 4x - 2x - 2 - 4 = 0 $

$ 2x^2 - 4x - 5 = 0 $

Let $ x_1 $ and $ x_2 $ be the roots of the quadratic equation above. By Vieta's formulas:

$ x_1 + x_2 = 4/2 = 2 $

The midpoint $ M(a, b) $ of $ AB $ on the x-axis is:

$ \frac{x_1 + x_2}{2} = 1 $

Now substitute $ x = -y - 1 $ into the circle equation:

$ (-y - 1)^2 + y^2 - 4(-y - 1) + 2y - 4 = 0 $

Simplify:

$ (y^2 + 2y + 1) + y^2 + 4y + 4 + 2y - 4 = 0 $

$ 2y^2 + 8y + 1 = 0 $

Let $ y_1 $ and $ y_2 $ be the roots of this quadratic equation. By Vieta's formulas:

$ y_1 + y_2 = -8/2 = -4 $

The midpoint $ M(a, b) $ on the y-axis is:

$ \frac{y_1 + y_2}{2} = -2 $

Thus, the midpoint coordinates are $ (1, -2) $.

Therefore, the values are:

$ a = 1, \quad b = -2 $

Calculating $ a - b $:

$ a - b = 1 - (-2) = 3 $

Equation of circle passes through the point of intersection of

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

and $ x^{2}+y^{2}-2 y=0 $ is

$ x^{2}+y^{2}-2 x-3+\lambda\left(x^{2}+y^{2}-2 y\right)=0 $

$ x^{2}(1+\lambda)+y^{2}(1+\lambda)-2 x-2 \lambda y-3=0 $

$ x^{2}+y^{2}-\frac{2 x}{1+\lambda}-\frac{\lambda}{1+\lambda} 2 y-\frac{3}{1+\lambda}=0 $

Centre $ \left(\frac{1}{1+\lambda}, \frac{\lambda}{1+\lambda}\right) $

Radius $ =\sqrt{\frac{1}{(1+\lambda)^{2}}+\frac{\lambda^{2}}{1+\lambda^{2}}+\frac{3}{1+\lambda}} $

$ \begin{aligned} & =\sqrt{\frac{1+\lambda^2+3+3 \lambda}{(1+\lambda)^2}}=\sqrt{\frac{\lambda^2+3 \lambda+4}{(\lambda+1)^2}} \end{aligned} $

$x+y+1=0$ touches the circle

$ \begin{aligned} & \frac{\frac{1}{1+\lambda}+\frac{\lambda}{1+\lambda}+1}{\sqrt{2}}=\frac{\sqrt{\lambda^2+3 \lambda+4}}{\lambda+1} \\\\ & \frac{1+\lambda}{1+\lambda}+1=\frac{\sqrt{\lambda^2+3 \lambda+4}}{\lambda+1}(\sqrt{2}) \\\\ & \sqrt{2}(\lambda+1)=\sqrt{\lambda^2+3 \lambda+4} \\\\ & 2 \lambda^2+2+4 \lambda=\lambda^2+3 \lambda+4 \\\\ & \lambda^2+\lambda-2=0 \\\\ & (\lambda+2)(\lambda-1)=0 \\\\ & \lambda=-2 \text { or } 1 \end{aligned} $

Take $\lambda=1$, we get

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

To find the center of circle $ S $, where the common chord of the given circles serves as its diameter, we first determine the equation of the common chord.

The equations of the two circles are:

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

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

To find the common chord, subtract the second circle's equation from the first:

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

Next, find the centers of the circles:

First circle, center $ C_1 = (1, -1) $

Second circle, center $ C_2 = (1, 1) $

The midpoint of $ C_1 $ and $ C_2 $ is calculated as follows:

$ \left(\frac{1 + 1}{2}, \frac{-1 + 1}{2}\right) = (1, 0) $

Thus, the center of circle $ S $ is at the intersection of the common chord $ y = -\frac{3}{4} $ and the line connecting $ C_1 $ and $ C_2 $ which is the diameter of $ S $. Hence, the center of circle $ S $ is:

$ (1, -\frac{3}{4}) $

Points of intersection of circles is

$ \begin{array}{l} \quad x^{2}+y^{2}+4 x-12=x^{2}+y^{2}+4 x-12 \\\\ \therefore \quad x=0 \end{array} $

Now, $ y= \pm 2 \sqrt{3} $

Now, area of rhombus

$ \begin{array}{l} =\frac{1}{2} \times d_{1} \times d_{2} \\\\ =\frac{1}{2} \times 4 \sqrt{3} \times 4 \\\\ =8 \sqrt{3} \text { sq units } \end{array} $