Circle

$ m=\left(\frac{2+4}{2}, \frac{5+3}{2}\right)=(3,4) $

and $m Q \leq 5$

So, $P$ should be in the shaded region thus, $A B:(y-3)=\frac{3-5}{4-2}(x-4)$

$ \begin{aligned} & \Rightarrow \quad y=-x+4+3 \\ & \Rightarrow \quad x+y-7=0 \end{aligned} $

and equation of the semi-circle

$ \begin{aligned} & (x-3)^2+(y-4)^2=(5)^2 \\ & \Rightarrow \quad x^2-6 x+9+y^2-8 y+16=25 \\ & \Rightarrow \quad x^2+y^2-6 x-8 y=0 \end{aligned} $

Hence, shaded region is

$ x^2+y^2-6 x-8 y \leq 0, x+y-7<0 $

$x^2+y^2-2 x-4 y-4=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ \begin{aligned} & c_1=(1,2) \\ & \text { and } r_1=\sqrt{(1)^2+(2)^2-(-4)}=3 \\ & \quad x^2+y^2+2 x+4 y-11=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & c_2=(-1,-2) \text { and } \\ & r_2=\sqrt{(-1)^2+(-2)^2-(-11)}=4 \\ & c_1 c_2=\sqrt{(-1-1)^2+(-2-2)^2} \\ & \quad=\sqrt{18} < r_1+r_2=7 \end{aligned} $

So, both circles intersect, each others.

Subracting Eq. (i) from Eq. (ii), we get $4 x+8 x-7=0$

Hence, intersects at the points lying on the line $4 x+8 y-7=0$

$ \begin{aligned} & \text { } x^2+y^2-6 x+4 y-12=0 \\ & \Rightarrow\left(x^2-6 x+9\right)+\left(y^2+4 y+4\right)=12+9+4 \\ & \Rightarrow(x-3)^2+(y+2)^2=(5)^2 \end{aligned} $

Centre $=(3,-2)$ and radius $=5$

The line $4 x-3 y+7=0$ touches the circle at the point ( $\alpha, \beta$ )

Foot of perpendicular from $(3,-2)$ to the line

$ \begin{aligned} & 4 x-3 y+7=0 \text { is } \\ & \frac{x-x_0}{a}=\frac{y-y_0}{b}=-\frac{a x_0+b y_0+c}{a^2+b^2} \\ & \Rightarrow \frac{x-3}{4}=\frac{y+2}{-3}=-\frac{4(3)-3(-2)+7}{3^2+4^2}=-1 \end{aligned} $

Thus, $\frac{x-3}{4}=-1 \Rightarrow x=-1$

and $\frac{y+2}{-3}=-1 \Rightarrow y=+1$

so, $(\alpha, \beta)=(-1,1)$

Hence, $\alpha+2 \beta=-1+2=1$

$ \begin{aligned} & \text { } x^2-4 x+y^2+12 y=216 \\ & \Rightarrow(x-2)^2+(y+6)^2=256 \\ & c_1=(2,-6), r_1=16 \\ & \text { and } x^2+6 x+y^2-12 y=-36 \\ & \Rightarrow(x+3)^2+(y-6)^2=9 \\ & \qquad \begin{aligned} \therefore c_2 & =(-3,6), r_2=3 \\ c_1 c_2 & =\sqrt{(2+3)^2+(-6-6)^2} \\ & =\sqrt{169}=13 \end{aligned} \end{aligned} $

and $\quad r_1-r_2=16-3=13$

$\because c_1 c_2=\left|r_1-r_2\right|$ the circle touches internally.

then, the slope of line joining centers

$ =\frac{6-(-6)}{3-2}=-12 / 5 $

$\therefore$ the slope of tangent

$ =\frac{-1}{(-12 / 5)}=5 / 12 $

The four given circles are

$ (x \pm r)^2+(y \pm r)^2=r^2 $

centers are $(r, r),(r,-r),(-r,-r),(-r, r)$ the circle touches all four given centered at the origin $(0,0)$, due to the symmetry.

One circle is internally tangent to the region by the four circles, with radius $\left(r_1\right)=r(\sqrt{2}-1)$

Other circles is externally tangent to the region, with radius $\left(r_2\right)=r(\sqrt{2}+1)$

$ \therefore r_1+r_2=r(\sqrt{2}-1+\sqrt{2}+1)=2 \sqrt{2} r $

Hence, $\frac{r_1+r_2}{r}=\frac{2 \sqrt{2} r}{r}=2 \sqrt{2}$

$\because x^2+y^2+x-3 y-10=0$ and $x^2+y^2+2 x-y-20=0$

So, common chord will be

$ \begin{aligned} & \left(x^2+y^2+x-3 y-10\right) -\left(x^2+y^2+2 x-y-20\right)=0 \\ & \Rightarrow x+2 y-10=0 \Rightarrow x=10-2 y \end{aligned} $

put the ' $x$ ' in first circle

$ \begin{aligned} & (10-2 y)^2+y^2+(10-2 y)-3 y-10=0 \\ & \Rightarrow 100+4 y^2-40 y+y^2-2 y+10 -3 y-10=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow y^2-9 y+20=0 \\ & \Rightarrow(y-5)(y-4)=0 \\ & \text { when } y=4 \Rightarrow x=2 \\ & \text { when } y=5 \Rightarrow x=0 \end{aligned} $

so, intersection points are $(2,4)$ and $(0,3)$ which are end-points of diameters.

So, equation of the circle

$ \begin{array}{ll} & (x-0)(x-2)+(y-4)(y-5)=0 \\ \Rightarrow \quad & x(x-2)+(y-4)(y-5)=0 \\ \Rightarrow \quad & x^2+y^2-2 x-9 y+20=0 \end{array} $

Comparing with

$ \begin{aligned} & x^2+y^2+\alpha x+\beta y+\gamma=0 \\ & \alpha=-2, \beta=-9 \text { and } \gamma=20 \end{aligned} $

Therefore, $\alpha+2 \beta+\gamma$

$ =-2-18+20=0 $

$\therefore \triangle A B C$ is right-angle triangle, right angle at $A$

$ \begin{array}{ll} \therefore & m_{A C} \cdot m_{A B}=-1 \\ & \frac{y-5}{x-4} \cdot \frac{y-2}{x-1}=-1 \\ \Rightarrow & (x-4)(x-1)+(y-5)(y-2)=0 \\ \Rightarrow & x^2+y^2-5 x-7 y+14=0 \end{array} $

$ \begin{aligned} &\text { Here, } r=\left|\frac{4 r+3 r-6}{5}\right|\\ &\begin{array}{lll} 5 r=7 r-6 & -5 r=7 r-6 \\ 6=2 r & \text { or } & r=\frac{6}{12} \\ r=3 & \text { or } & r=\frac{1}{2} \end{array} \end{aligned} $

∴ Centre $(3,3)$ or $\left(\frac{1}{2}, \frac{1}{2}\right)$

∵ Circle lies below the line

∴ Centre $\left(\frac{1}{2}, \frac{1}{2}\right), r=\frac{1}{2}$

∴ Equation of circle,

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

Equation of circle passing through $x^2+y^2-a^2=0$ and $x \cos \alpha+y \sin \alpha-p=0$, is

$ \begin{aligned} & S+\lambda L=0 \\ & \left(x^2+y^2-a^2\right)+\lambda(x \cos \alpha+y \sin \alpha-p)=0 \\ & x^2+y^2+x(\lambda \cos \alpha)+y(\lambda \sin \alpha)-a^2-\lambda p \\ & =0 \\ & \therefore r=\sqrt{\frac{\lambda^2 \cos ^2 \alpha}{4}+\frac{\lambda^2 \sin ^2 \alpha}{4}+a^2+\lambda p} \\ & \Rightarrow r=\sqrt{\frac{\lambda^2}{4}+\lambda p+a^2} \\ & =\sqrt{\left(\frac{\lambda}{2}+p\right)^2+a^2-p^2} \\ & \quad \min r \Rightarrow \frac{\lambda}{2}+p=0 \\ & \Rightarrow \quad \lambda=-2 p \end{aligned} $

$ \begin{aligned} & L_1: 3 x-4 y+4=0 \\ & \quad L_2: 3 x-4 y-\frac{7}{2}=0 \end{aligned} $

$ \begin{aligned} &\because L_1 \| L_2\\ &\therefore 2 r=\text { Distance between } L_1 \text { and } L_2\\ &\begin{aligned} = & \left|\frac{4+\frac{7}{2}}{\sqrt{3^2+4^2}}\right|=\left|\frac{15}{2 \times 5}\right| \\ 2 r & =\frac{3}{2} \Rightarrow r=\frac{3}{4} \\ \therefore \text { Area }= & \pi r^2=\frac{9 \pi}{16} \end{aligned} \end{aligned} $

Let equation of circle be

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

It passes through $(2,0)$

$ \begin{aligned} \therefore \quad 4+4 g+c & =0 \\ x_{\text {intercept }} & =5 \\ 2 \sqrt{g^2-c} & =5 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad g^2+4+4 g=\frac{25}{4} \\ & \Rightarrow \quad 4 g^2+16+16 g-25=0 \\ & \Rightarrow \quad 4 g^2+16 g-9=0 \\ & \quad g=\frac{-9}{2}, \frac{1}{2} \end{aligned} $

∵ Centre lies in 1st quadrant $\therefore g<0$

$ \begin{array}{ll} \therefore & g=-\frac{9}{2} \\ \Rightarrow & c=-4-4 g=14 \end{array} $

Equation of circle,

$ x^2+y^2-9 x+2 f y+14=0 $

∵ Centre lies in Ist quadrant

∴ Equation $x^2+y^2-9 x-2 k y+14=0$,

$k \in R^{+}$

Let the centroid of the triangle be at $(h, k)$.

Step 1: Finding the Centroid Coordinates

The centroid's $x$-coordinate ($h$) is the average of all three $x$-coordinates: $ h = \frac{\cos \alpha + \sin \alpha + 1}{3} $ So, $ 3h - 1 = \cos \alpha + \sin \alpha $

The centroid's $y$-coordinate ($k$) is the average of all three $y$-coordinates: $ k = \frac{\sin \alpha - \cos \alpha + 2}{3} $ So, $ 3k - 2 = \sin \alpha - \cos \alpha $

Step 2: Squaring Both Sides

Square both sides of each equation to remove the trigonometric functions:

$ (3h - 1)^2 = (\cos \alpha + \sin \alpha)^2 \quad \cdots (i) $

$ (3k - 2)^2 = (\sin \alpha - \cos \alpha)^2 \quad \cdots (ii) $

Step 3: Add the Two Equations

Add equations $(i)$ and $(ii)$: $ (3h - 1)^2 + (3k - 2)^2 = (\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2 $

Notice that $(\cos \alpha + \sin \alpha)^2 + (\sin \alpha - \cos \alpha)^2 = 2\sin^2 \alpha + 2\cos^2 \alpha = 2 $

So, $ (3h - 1)^2 + (3k - 2)^2 = 2 $

Step 4: Locus Equation

Replace $h$ by $x$ and $k$ by $y$ because the locus is about every possible point $(x, y)$: $(3x - 1)^2 + (3y - 2)^2 - 2 = 0$

Step 5: Expand to Standard Form

Expanding and simplifying:

$ \begin{aligned} & (3x - 1)^2 + (3y - 2)^2 - 2 = 0 \\ & 9x^2 - 6x + 1 + 9y^2 - 12y + 4 - 2 = 0 \\ & 9x^2 + 9y^2 - 6x - 12y + 3 = 0 \\ & 3(x^2 + y^2) - 2x - 4y + 1 = 0 \end{aligned} $

$ \because A B \text { will be the diameter } $

$ \begin{aligned} &\text { ∴ Let equation of circle }\\ &x^2+y^2+2 g x+2 f y=0 \end{aligned} $

( $a, 0$ ) lies on the circle

$ \therefore \quad a^2+2 g a=0 \Rightarrow a=-2 g $

$(0, b)$ lies on the circle

$ \therefore \quad b^2+2 f b \Rightarrow b=-2 f $

∴ Equation of line $A B$

$ \begin{aligned} & \frac{x}{-2 g}+\frac{y}{-2 f}=1\left[\because\left(x_1, y_1\right) \text { lies on } L\right] \\ \Rightarrow \quad & \frac{x}{-g}+\frac{y}{-f}=2 \Rightarrow \frac{x_1}{-g}+\frac{y_1}{-f}=2 \end{aligned} $

∴ Locus is centre $(-g,-f)$ is $\frac{x_1}{x}+\frac{y_1}{y}=2$

Given, $C_1: x^2+y^2=3$,

$ O_1(0,0), r_1=\sqrt{3} $

and $C_2: x^2+y^2-2 x+4 y+4=0$

$ O_2(1,-2), r_2=1 $

Point $P$ divides $O_1 O_2$ externally in $r_1: r_2$

$ \begin{aligned} \therefore & \alpha=\frac{0 \times \sqrt{3}-1(1)}{\sqrt{3}+1}=\frac{-1}{\sqrt{3}+1} \text { and } \\ & \beta=\frac{0 \times \sqrt{3}-(-2)}{\sqrt{3}+1}=\frac{2}{\sqrt{3}+1} \\ \Rightarrow \quad & \frac{\beta}{\alpha}=-2 \end{aligned} $

$ \begin{aligned} &\text { Given, }|x-2|+|y-3|=4\\ &\begin{array}{ll} \therefore & x-2+y-3=4 \\ \Rightarrow & x+y=9 \text { and } x-2-y+3=4 \\ \Rightarrow & x-y=3 \text { and }-x+2+y-3=4 \\ \Rightarrow & x-y=-5 \text { and }-x+2-y+3=4 \\ \Rightarrow & x+y=1 \end{array} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \therefore \text { Centre }\left(\frac{4+0}{2}, \frac{5+1}{2}\right)=(2,3) \\ & \text { and } r=\frac{\sqrt{4^2+4^2}}{2}=2 \sqrt{2} \end{aligned}\\ &\text { ∴ Equation of circle : }\\ &\begin{aligned} & (x-2)^2+(y-3)^2=(2 \sqrt{2})^2 \\ \Rightarrow \quad & x^2+y^2-4 x-6 y+5=0 \end{aligned} \end{aligned} $

$ \begin{gathered} A B=\sqrt{1^2+9}=\sqrt{10} \\ A M=\frac{\sqrt{10}}{2} \end{gathered} $

$ \therefore O A=r=\sqrt{\frac{10}{4}+\frac{10}{4}}=\sqrt{\frac{20}{4}}=\sqrt{5} $

Let $O(\alpha, \beta)$

$ \begin{array}{ll} & O A=O B \\ & (\alpha-1)^2+(\beta-2)^2=(\alpha-2)^2+(\beta+1)^2 \\ \Rightarrow & -2 \alpha+1-4 \beta+4=-4 \alpha+4+2 \beta-1 \\ \Rightarrow & 2 \alpha-6 \beta=0 \\ \Rightarrow & \quad \alpha=3 \beta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ \because & m_{O A} \cdot m_{O B}=-1 \\ \Rightarrow & \quad \frac{\beta-2}{\alpha-1} \cdot \frac{\beta+1}{\alpha-2}=-1 \\ \Rightarrow & (\beta-2)(\beta+1)+(\alpha-1)(\alpha-2)=0 \\ \Rightarrow & \beta^2-\beta-2+\alpha^2-3 \alpha+2=0 \\ \Rightarrow & \beta^2-\beta+9 \beta^2-9 \beta=0 \\ \Rightarrow & 10 \beta^2=10 \beta \\ \Rightarrow & \quad \beta=0,1 \\ & \quad \alpha=0,3 \end{array} $

∴ Centre $(0,0)$ or $(3,1)$

∴ Equation of circle

$ \begin{aligned} & x^2+y^2=5 \text { or }(x-3)^2+(y-1)^2=5 \\ & \Rightarrow x^2+y^2-6 x-2 y+5=0 \end{aligned} $

Given,

$ \begin{aligned} & \quad S_1: 4(x-1)^2+4(y-1)^2=1 \\ & \Rightarrow(x-1)^2+(y-1)^2=\frac{1}{4} \\ & \qquad O_1(1,1), r_1=\frac{1}{2} \\ & S_2: 4(x+1)^2+4(y-1)^2=1 \\ & \Rightarrow(x+1)^2+(y-1)^2=\frac{1}{4} \\ & O_2(-1,1), r_2=1 / 2 \\ & \text { and } S_3: 4(x+1)^2+4(y+1)^2=1 \\ & \Rightarrow(x+1)^2+(y+1)^2=\frac{1}{4} \\ & O_3(-1,-1), r_3=\frac{1}{2} \\ & \because s_1-s_2=0 \\ & \Rightarrow-2 x-2 y+2 x-2 y=0 \\ & \quad x=0 \\ & \quad S_2-s_3=0 \\ & \Rightarrow \quad 2 y=0 \\ & \Rightarrow \quad y=0 \end{aligned} $

∴ Centre of required circle $=(0,0)$

$ \text { radius }=L=\sqrt{s_1} $

$ \begin{aligned} &=\sqrt{1+1-\frac{1}{4}}=\sqrt{\frac{7}{4}}\\ &\text { ∴ Equation of circle }\\ &\begin{aligned} & x^2+y^2=\frac{7}{4} \\ \Rightarrow & 4 x^2+4 y^2=7 \end{aligned} \end{aligned} $

Let the coordinates of the centroid be $G(x, y)$ of the $\triangle A P Q$ is

$ x=\frac{x_A+x_P+x_Q}{3}, y=\frac{y_A+y_P+y_Q}{3} $

Given the points $A(a, 0), P(a \cos \theta, a \sin \theta)$ and $Q(b \sin \theta,-b \cos \theta)$

The coordinates of the centroid are

$ \begin{aligned} & x=\frac{a+a \cos \theta+b \sin \theta}{3} \\ & y=\frac{0+a \sin \theta-b \cos \theta}{3} \end{aligned} $

Let $x=h$ and $y=k$, then,

$ \begin{aligned} & h=\frac{a+a \cos \theta+b \sin \theta}{3}, k=\frac{a \sin \theta-b \cos \theta}{3} \\ & \Rightarrow 3 h-a=a \cos \theta+b \sin \theta \\ & 3 k=a \sin \theta-b \cos \theta \end{aligned} $

Squaring and adding these equations

$ \begin{aligned} (3 h-a)^2+(3 k)^2=(a \cos \theta & +b \sin \theta)^2 +(a \sin \theta-b \cos \theta)^2 \end{aligned} $

$ \begin{aligned} = & a^2 \cos ^2 \theta+b^2 \sin ^2 \theta+2 a b \cos \theta \sin \theta +a^2 \sin ^2 \theta+b^2 \cos ^2 \theta-2 a b \cos \theta \sin \theta \\ = & a^2+b^2 \end{aligned} $

Replacing $h$ with $x$ and $k$ with $y$, we get the locus of the centroid as

$ \begin{aligned} & (3 x-a)^2+(3 y)^2=a^2+b^2 \\ \Rightarrow \quad & \left(x-\frac{a}{3}\right)^2+y^2=\frac{a^2+b^2}{9} \\ \Rightarrow \quad & \left(x-\frac{a}{3}\right)^2+y^2=\left(\frac{\sqrt{a^2+b^2}}{3}\right)^2 \end{aligned} $

So, the locus of the centroid is

$ \left(x-\frac{a}{3}\right)^2+y^2=\left(\frac{\sqrt{a^2+b^2}}{3}\right)^2 $

It is a circle with centre at $\left(\frac{a}{3}, 0\right)$ and radius $\frac{\sqrt{a^2+b^2}}{3}$

Given, lines are $x+2 y-2=0$ and

$ 2 x+3 y-1=0 $

Solving these two lines, we get

$ x=-4 \text { and } y=3 $

Thus, the center of circle is $(-4,3)$.

Since, the general eq. of circle is $(x-h)^2+(y-k)^2=r^2$, where $(h, k)$ is the center and $r$ is the radius.

$ \begin{aligned} & \text { So, }(x-(-4))^2+(y-3)^2=r^2 \\ & \Rightarrow \quad(x+4)^2+(y-3)^2=r^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

But, the circle passes through the point $(8,8)$,

$ \begin{aligned} & \text { So, }(8+4)^2+(8-3)^2=r^2 \\ & 12^2+5^2=r^2 \\ &\,\,\,\,\,\, \quad 169=r^2 \\ & \Rightarrow \quad r=13 \end{aligned} $

Now, from Eq. (i), we get

$ \begin{aligned} \quad(x+4)^2+(y-3)^2=r^2 & =169 \\ \Rightarrow \quad x^2+8 x+16+y^2-6 y+9 & =169 \\ \Rightarrow \quad x^2+y^2+8 x-6 y-144 & =0 \end{aligned} $

Comparing with the given equation

$ x^2+y^2+p x+q x+r=0 $

We get $p=8, q=-6, r=-144$

$ \begin{aligned} \therefore \quad p^2+q^2+r & =8^2+(-6)^2+(-144) \\ & =64+36-144 \\ & =-44 \end{aligned} $

Given circle is

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

So, center $=(-g,-f)=\left(\frac{-2}{2}, \frac{4}{2}\right)$

$ =(-1,2) \text { and } c=3 $

Also, the equation of the polar of a point $\left(x_1, y_1\right)$ w.r.t the circle $x^2+y^2+2 g x+2 f y+c=0$ is

$ x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0 $

$ \begin{aligned} &\begin{aligned} & \Rightarrow x x_1+y y_1-\left(x+x_1\right)+2\left(y+y_1\right)+3=0 \\ & \Rightarrow x x_1+y y_1-x-x_1+2 y+2 y_1+3=0 \\ & \Rightarrow\left(x_1-1\right) x+\left(y_1+2\right) y-x_1+2 y_1+3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { Given, polar equation is }\\ &2 x-3 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

By comparing the coefficients of two equations, we get

$ \begin{aligned} \frac{x_1-1}{2} & =\frac{y_1+2}{-3} \\ & =\frac{-x_1+2 y_1+3}{1} \\ \Rightarrow \quad \frac{x_1-1}{2} & =\frac{y_1+2}{-3} \text { and } \\ \frac{x_1-1}{2} & =\frac{-x_1+2 y_1+3}{1} \\ \Rightarrow-3 x_1+3 & =2 y_1+4 \text { and } x_1-1 \\ & =-2 x_1+4 y_1+6 \\ \Rightarrow 3 x_1+2 y_1 & =-1 \text { and } 3 x_1-4 y_1-7=0 \end{aligned} $

By solving these two equations, we get $x_1=\frac{5}{9}$ and $y_1=\frac{-4}{3}$

$ \text { } \begin{aligned}Now,\,\, 3 x_1-y_1 & =3\left(\frac{5}{9}\right)-\left(\frac{-4}{3}\right) \\ & =\frac{5}{3}+\frac{4}{3}=\frac{9}{3}=3 \\ \therefore \quad 3 x_1-y_1 & =3 \end{aligned} $

Given,

$ \begin{aligned} & \quad s \equiv x^2+y^2+2 g x+2 f y+c=0 \\ & \text { And } s^{\prime} \equiv x^2+y^2-6 x+6 y+2=0 \\ & \Rightarrow \quad\left(x^2-6 x\right)+\left(y^2+6 y\right)=-2 \\ & \Rightarrow \quad\left(x^2-6 x+9\right)+\left(y^2+6 y+9\right) \\ & \quad=-2+9+9 \\ & \Rightarrow \quad(x-3)^2+(y+3)^2=16 \end{aligned} $

Thus, the center of circle $S^{\prime}$ is $(3,-3)$ and radius, $r^{\prime}=4$

$ \begin{aligned} & \text { Now, } S \equiv x^2+y^2+2 g x+2 f y+c=0 \\ & \Rightarrow \quad(x+g)^2+(y+f)^2=g^2+f^2-c \end{aligned} $

So, the center of circle $S$ is $(-g,-f)$ and radius,

$ r=\sqrt{g^2+f^2-c} $

Since, radius of circle $S$ is 1 .

$ \begin{array}{lr} \therefore & \sqrt{g^2+f^2-c}=1 \\ \Rightarrow & g^2+f^2-c=1 \end{array} $

Also, two circle touch externally,

So, distance between $(-g,-f)$ and

$ \begin{aligned} & (3,-3)=r+r^{\prime} \\ & \Rightarrow \quad \sqrt{(-g-3)^2+(-f+3)^2}=4+1 \\ & \Rightarrow \quad(g+3)^2+(f-3)^2=25 \\ & \Rightarrow \quad g^2+6 g+9+f^2-6 f+9=25 \\ & \Rightarrow \quad g^2+f^2+6 g-6 f+18-25=0 \\ & \Rightarrow \quad g^2+f^2+6 g-6 f=7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Since, point $(-1,-3)$ lies on both circles, it satisfies both circle.

$\therefore$ For circles

$ \begin{aligned} & (-1)^2+(-3)^2+2 g(-1)+2 f(-3)+c=0 \\ & \Rightarrow \quad 1+9-2 g-6 f+c=0 \\ & \Rightarrow \quad-2 g-6 f+c=-10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

For circle $S^{\prime}$,

$ \begin{aligned} & (-1)^2+(-3)^2-6(-1)+6(-3)+2=0 \\ & \Rightarrow \quad 1+9+6-18+2=0 \\ & \Rightarrow \quad 0=0 \end{aligned} $

Now, the slope of line joining $(-g,-f)$ and $(-1,-3)$ is equal to slope of line joining $(-1,-3)$ and $(3,-3)$

$ \begin{aligned} & & (-f+3) & =\left(\frac{-3+3}{3+1}\right)(-g+1) \\ \Rightarrow & & (-f+3) & =0 \\ \Rightarrow & & f & =3\end{aligned} $

Put $f=3$ into Eq. (i), we get

$ \begin{aligned} & & & g^2+9+6 g-6(3) \\ \Rightarrow & & g^2+9+6 g-18 & =7 \\ \Rightarrow & & g^2+6 g-16 & =0 \\ \Rightarrow & & (g+8)(g-2) & =0 \\ \Rightarrow & & g & =-8 \text { or } g=2 (not \,\,possible)\end{aligned} $

Now, put $g=2, f=3$ into Eq. (ii), we get

$ \begin{aligned} & -2(2)-6(3)+c=-10 \\ \Rightarrow \quad & c=-10+4+18=12 \end{aligned} $

Now, $g+f+c=2+3+12=17$

$ \begin{aligned} & S \equiv x^2+y^2-6 x+8 y+k=0 \\ & \Rightarrow(x-3)^2+(y+4)^2=25-k \end{aligned} $

Thus, the center of the circle $C$ is at $(3,-4)=\left(x_1, y_1\right)$.

Let point $P$ as $\left(x_2, y_2\right)$ and point $Q$ is the point that divides the line segment $C P$ in the ratio $-1: 2$

Also, $3 x+4 y-43=0$

$ \Rightarrow \quad y=\frac{-3}{4} x+\frac{43}{3} $

So, slope of tangent line $=\frac{-3}{4}$

And slope of line $C P=-\left(\frac{-4}{3}\right)=\frac{4}{3}$

Thus, equation of line $C P$ is

$ \begin{array}{lrl} \Rightarrow & y-(-4) & =\frac{4}{3}(x-3) \\ \Rightarrow & y+4 & =\frac{4}{3}(x-3) \\ \Rightarrow & y & =\frac{4}{3} x-8 \end{array} $

Since, $P$ lies on both the lines $3 x+4 y-43=0$ and the line $C P$,

$ \begin{array}{ll} \therefore & 3 x+4\left(\frac{4}{3} x-8\right)-43=0 \\ \Rightarrow & 3 x+\frac{16}{3} x-32-43=0 \end{array} $

$ \begin{aligned} & \Rightarrow \quad \frac{25}{3} x=75 \\ & \Rightarrow \quad x=9 \\ & \text { Now, } y=\frac{4}{3} x-8=\frac{4}{3}(9)-8 \\ & =12-8=4 \end{aligned} $

So, $\quad P=(9,4)$

Using section formula, we find the coordinates of $Q(x, y)$.

$ \begin{aligned} & x=\frac{2 x_1-x_2}{2-1}=\frac{2(3)-9}{1}=-3 \\ & y=\frac{2 y_1-y_2}{2-1}=\frac{2(-4)-4}{1}=-12 \end{aligned} $

So, $Q=(x, y)=(-3,-12)$

The power of point $Q$ w.r.t to circle, $x^2+y^2-6 x+8 y+k=0$ is

$ \begin{aligned} (-3)^2 & +(-12)^2-6(-3)+8(-12)+k \\ & =9+144+18-96+k \\ & =75+k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, the line $3 x+4 y-43=0$ is tangent to the circle, the radius of the circle is the distance from the center $C(3,-4)$ to the line $3 x+4 y-43=0$

$ \begin{aligned}So,\,r & =\frac{\left|A x_0+B y_0+C\right|}{\sqrt{A^2+B^2}} \\ & =\frac{|3(3)+4(-4)-43|}{\sqrt{3^2+4^2}} \\ & =\frac{|9-16-43|}{5}=\frac{50}{5}=10 \end{aligned} $

Now, $r=\sqrt{25-k}$

$ \begin{aligned} & \Rightarrow \quad 10=\sqrt{25-k} \Rightarrow 25-k=100 \\ & \Rightarrow \quad k=-75 \end{aligned} $

Put $k=-75$ in Eq. (i), we get The power of $Q$ w.r.t the circle

$ =75+k=75-75=0 $

$ \begin{array}{r} s_1: x^2+y^2+2 g x+2 f y+c=0 \\ s_2: 2 x^2+2 y^2+3 x+8 y+2 c=0 \\ \Rightarrow \quad x^2+y^2+\frac{3}{2} x+4 y+c=0 \\ s_3: x^2+y^2+2 x+2 y+1=0 \end{array} $

Radical axis $s_1-s_2=0$

$ \begin{array}{ll} \Rightarrow & \left(x^2+y^2+2 g x+2 f y+c\right) \\ & -\left(x^2+y^2+\frac{3}{2} x+4 y+c\right)+0 \\ \Rightarrow & 2 g x+2 f y-\frac{3 x}{2}-4 y=0 \\ \Rightarrow & \left(2 g-\frac{3}{2}\right) x+(2 f-4) y=0 \\ \Rightarrow & (4 g-3) x+(4 f-8) y=0 \end{array} $

The radical axis touches the circle $s_3$, where radius of $s_3$.

$=\sqrt{(-1)^2+(-1)^2-1}=1$ and center is $(-1,-1)$

The perpendicular distance from the center of $s_3$ to the radical axis $=$ radius of $s_3$

$ \begin{aligned} & \text { } \frac{|(4 g-3) \cdot(-1)+(4 f-8)(-1)|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=1 \\ & \Rightarrow \quad \frac{|-4 g+3-4 f+8|}{\sqrt{(4 g-3)^2+(4 f-8)^2}}=1 \\ & \Rightarrow \quad|-4 g-4 f+11| \\ & \quad=\sqrt{(4 g-3)^2+(4 f-8)^2} \\ & \Rightarrow \quad(-4 g-4 f+11)^2 \\ & \quad=(4 g-3)^2+(4 f-8)^2 \\ & \Rightarrow \quad 16 g^2+16 f^2+121+32 g f-88 g-88 f \\ & \quad=16 g^2-24 g+9+16 f^2-64 f+64 \\ & \Rightarrow \quad 32 g f-64 g-24 f+48=0 \\ & \Rightarrow \quad 8 g f-16 g-6 f+12=0 \\ & \Rightarrow \quad 8 g(f-2)-6(f-2)=0 \\ & \Rightarrow \quad(8 g-6)(f-2)=0 \\ & \Rightarrow \quad(4 g-3)(f-2)=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 4 g-3=0 \text { or } f-2=0 \\ & \Rightarrow \quad g=\frac{3}{4} \text { or } f=2 \end{aligned} $

Either $g=\frac{3}{4}$ or $f=2$ is the required condition.

$\because x^2+y^2-2 x-4 y-20=0$

After rotation of axes through $\theta=\pi / 4$, it transforms to $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ in the given equation

$ a=1, h=0, b=1, g=-1, f=-2, c=-20 $

After rotation of $\theta=\pi / 4$

$ \begin{aligned} a^{\prime} & =a \cos ^2 \theta+b \sin ^2 \theta+2 h \cos \theta \sin \theta \\ & =1\left(\frac{1}{\sqrt{2}}\right)^2+1\left(\frac{1}{\sqrt{2}}\right)^2+0=1 \\ b^{\prime}= & a \sin ^2 \theta+b \cos ^2 \theta-2 h \sin \theta \cos \theta \\ = & 1\left(\frac{1}{\sqrt{2}}\right)^2+1\left(\frac{1}{\sqrt{2}}\right)^2-0=1 \end{aligned} $

also,

$ \begin{aligned} & h^{\prime}=(b-a) \sin \theta \cos \theta+h\left(\cos ^2 \theta-\sin ^2 \theta\right) \\ & =0+0=0 \end{aligned} $

Similarly, $x^{\prime}=x \cos \theta-y \sin \theta=\frac{1}{\sqrt{2}}(x-y)$

$ \begin{aligned} & =\frac{-6 x-2 y}{\sqrt{2}} \\ & =-3 \sqrt{2} x-\sqrt{2} y=2 g x+2 f y \end{aligned} $

then $g=-3 / \sqrt{2}, f=-1 / \sqrt{2}, c=-20$

So,

$ \begin{aligned} & \left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 1 & 0 & -3 / \sqrt{2} \\ 0 & 1 & -1 / \sqrt{2} \\ -3 / \sqrt{2} & -1 / \sqrt{2} & -20 \end{array}\right| \\ & \quad=1\left(-20-\frac{1}{2}\right)-\frac{3}{\sqrt{2}}\left(+\frac{3}{\sqrt{2}}\right) \\ & \quad=-41 / 2-9 / 2=-25 \end{aligned} $

Given equation of circle

$ x^2+y^2+5 k x+2 y+k=0 $

and $2 x^2+2 y^2+2 k x+3 y-1=0$

Equation of common chord

$ 8 k x+y+2 k+1=0 $

Since, this chord is coincide with

$ \begin{aligned} & 4 x+5 y-k=0 \\ & \frac{8 k}{4}=\frac{1}{5}=\frac{2 k+1}{-k} \end{aligned} $

Solving $\frac{8 k}{4}=\frac{1}{5}$

we get $k=\frac{1}{10}$

Solving $8 \frac{k}{4}=\frac{2 k+1}{-k} \Rightarrow 2 k^2+2 k+1=0$

No solution

$\therefore$ No values of $k$ possible

$ \begin{aligned} & \text { } c_1: x^2+y^2-2 x+4 y+1=0 \\ & \Rightarrow(x-1)^2+(y+2)^2=4 \\ & \text { Centre }\left(c_1\right)=(1,-2), r_1=2 \\ & \text { and } c_2: x^2+y^2-4 x-2 y+4=0 \\ & \Rightarrow(x-2)^2+(y-1)^2=1 \\ & \text { Centre }\left(c_2\right)=(2,1), r_2=1 \\ & c_1 c_2=d=\sqrt{(2-1)^2+\left(1+2^2\right.}=\sqrt{10} \end{aligned} $

Let the slope of direct common tangent be $m$ the general line is

$ y=m x+c $

this line has perpendicular distance $r_1$ from $c_1$ and $r_2$ from $c_2$

$ \begin{aligned} & \therefore\left|\frac{m(1)-(-2)+c}{\sqrt{1+m^2}}\right|=2 \\ & \Rightarrow|m+2+c|=2 \sqrt{1+m^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and }\left|\frac{m(2)-1+c}{\sqrt{1+m^2}}\right|=1 \\ & \Rightarrow|2 m+c-1|=\sqrt{1+m^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Put the value of ' $c$ ' from Eq. (i) to Eq. (ii)

$ 2 m-1+\left(-m-2+2 \sqrt{m^2+1}\right)=\sqrt{m^2+1} $

$ \begin{aligned} &\Rightarrow m=3-\sqrt{m^2+1}\\ &\text { Squaring both sides, we get }\\ &\begin{array}{rlrl} & m^2= & m^2+10 & -6 \sqrt{m^2+1} \\ \Rightarrow & & \sqrt{m^2+1} & =5 / 3 \\ \Rightarrow & & m^2+1 & =\frac{25}{9} \\ \Rightarrow & & m^2=16 / 9 \\ \Rightarrow & & m=4 / 3 \end{array} \end{aligned} $

We are told that $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle

$ x^2+y^2 = 25. $

Step 1. Condition for conjugate points

If $(x_1, y_1)$ and $(x_2, y_2)$ are conjugate points with respect to the circle $x^2+y^2=r^2$,

then the polar of one point passes through the other.

Equation of the polar of $(x_1, y_1)$ w.r.t. $x^2+y^2=r^2$ is:

$ x_1x + y_1y = r^2 $

Step 2. The point $(1,a)$

For the point $(1, a)$, the polar is:

$ 1 \cdot x + a \cdot y = 25 $

That is:

$ x + a y = 25. $

Step 3. The point $(b,2)$ lies on this polar

Since the points $(1,a)$ and $(b,2)$ are conjugate, the point $(b,2)$ lies on the polar of $(1,a)$.

So:

$ b + a(2) = 25 $

$ b + 2a = 25 \quad \text{(1)} $

Step 4. Symmetry of conjugate points

Similarly, the polar of $(b,2)$ is:

$ b x + 2 y = 25. $

And $(1,a)$ must lie on this polar:

$ b(1) + 2(a) = 25 $

Wait—that would give same, yes so far matches (1). Let's check the other way:

Substitute $ (x,y) = (1,a) $:

$ b \cdot 1 + 2 a = 25 $

This is the same as (1). So it’s consistent.

Step 5. We need $4a + 2b = ?$

From (1): $b + 2a = 25.$

Multiply both sides by 2:

$ 2b + 4a = 50 $

So,

$ \boxed{4a + 2b = 50.} $

✅ Final Answer: 50

Option B

Let $(h, k)$ be the pole of the line

$ x+2 b y-5=0 $

with respect to the circle

$ \begin{array}{ll} s \equiv x^2+y^2-4 x-6 y+4=0 \\ \therefore \quad h x+k y-2(x+h)-3(y+k)+4=0 \\ (h-2) x+(k-3) y-2 h-3 k+4=0 \end{array} $

is coincide with $x+2 b y-5=0$

$ \begin{aligned} & \frac{h-2}{1}=\frac{k-3}{2 b}=\frac{2 h+3 k-4}{5} \\ \Rightarrow & \frac{h-2}{1}=\frac{k-3}{2 b} \end{aligned} $

$ 2 b h-k=4 b-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\frac{h-2}{1}=\frac{2 h+3 k-4}{5}$

$ h-k=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$ h=\frac{4 b-5}{2 b-1}, k=\frac{-3}{2 b-1} $

$(h, k)$ lie on line $x+b y+1=0$

$ \begin{aligned} &\begin{aligned} & & \frac{4 b-5}{2 b-1}-\frac{3 b}{2 b-1}+1 & =0 \\ & & 4 b-5-3 b+2 b-1 & =0 \\ & & & =2 \end{aligned}\\ &\therefore \text { Pole of the line from pole }(2,-2) \text { is }\\ &\begin{aligned} & (2-2) x+(-2-3) y-2(2)-3(-2)+4=0 \\ & \Rightarrow \quad-5 y+6=0 \\ & \Rightarrow \quad 5 y-6=0 \end{aligned} \end{aligned} $

The given equations are

$ \begin{aligned} & s: x^2+y^2+4 x+7=0 \\ & s^{\prime}: 2 x^2+2 y^2+3 x+5 y+9=0 \\ & \Rightarrow x^2+y^2+\frac{3}{2} x+\frac{5}{2} y+\frac{9}{2}=0 \\ & s^{\prime \prime}: x^2+y^2+y=0 \end{aligned} $

Radical axis of $s$ and $s^{\prime}: s-s^{\prime}=0$

$ \Rightarrow \quad 4 x-\frac{3}{2} x-\frac{5}{2} y+7-\frac{9}{2}=0 $

$ \Rightarrow \quad \frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0 $

$ \Rightarrow \quad x-y+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

Radical axis of $s$ and $s^{\prime \prime}: s-s^{\prime \prime}=0$

$ \Rightarrow \quad 4 x-y+7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we have

$ x=-2, y=-1 $

$\therefore$ Radical centre is $P(-2,-1)$

Length of tangent from $P$ to $s^{\prime}$

$ \begin{aligned} & =\sqrt{(-2)^2+(-1)^2+\frac{3}{2}(-2)+\frac{5}{2}(-1)+\frac{9}{2}} \\ & =\sqrt{4+1-3+2}=\sqrt{4}=2 \end{aligned} $

Substitute $x=X+2$ and $y=Y-2$ into the original equation, $x^2+y^2=4$ we get;

$ \begin{aligned} & (X+2)^2+(Y-2)^2=4 \\ \Rightarrow & X^2+4+4 X+Y^2+4-4 Y=4 \\ \Rightarrow & X^2+Y^2+4 X-4 Y+4=0 \\ \therefore & a+b+c=4-4+4=4 \end{aligned} $

Given, point $P(-4,0)$

Equation of circle

$ x^2+y^2-4 x-6 y-12=0 $

Centre $(2,3)$

Centre of circle passing through $P, A, B$ is mid point of $O P$

$ \begin{array}{ll} \therefore\left(\frac{-4+2}{2}, \frac{0+3}{2}\right)=\left(-1, \frac{3}{2}\right) \\ \therefore (g, f)=\left(1, \frac{-3}{2}\right) \end{array} $

Equation of polar is $T=0$

$ \begin{aligned} & \text { i.e., } \alpha x+y(-1)-4 \cdot \frac{1}{2}(x+\alpha) \\ & \quad-6 \cdot \frac{1}{2}(y+(-1))-12=0 \\ & \Rightarrow \quad \alpha x-y-2 x-2 \alpha-3 y+3-12=0 \\ & \Rightarrow \quad(\alpha-2 x-4 y-2 \alpha-9=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

But equation of polar is $y=\beta$ given

So, coefficient of $x$ must be zero.

i.e. $\alpha-2=0 \Rightarrow \alpha=2$

Thus, $0-4 y-2 \times(2)-9=0\,\,\,\,(from Eq. (i))$

$ \begin{aligned} & \Rightarrow \quad-4 y=13 \\ & \Rightarrow \quad y=\frac{-13}{4}=\beta \,\,\,(given)\\ & \therefore 4(\alpha+\beta)=4\left(2-\frac{13}{4}\right)=-5 \end{aligned} $

Here centre of circle $=C=(2,3)$ let $P \equiv(-1,-1)$ and $C P=5$

Let radius $=r$

$ \begin{aligned} & \therefore \sin \frac{\theta}{2}=\frac{r}{C P} \\ & \Rightarrow \quad r^2=(5)^2 \sin ^2 \frac{\theta}{2} \\ & \because \cos \theta=\frac{-7}{25} \,\,\,(given)\end{aligned} $

$ \begin{aligned} & 1-2 \sin ^2 \frac{\theta}{2}=\frac{-7}{25} \Rightarrow 2 \sin ^2 \frac{\theta}{2}=1+\frac{7}{25} \\ & \Rightarrow \quad \sin ^2 \frac{\theta}{2}=\frac{16}{25}=25 \cdot \frac{16}{25}=16 \\ & \therefore \quad r=4\,\,\,\, (\because r>0)\end{aligned} $

$ \begin{aligned} &\text { As we know that, power of a point }\\ &\begin{aligned} & \text { W.r.t a circle }=\left(\sqrt{S_1}\right)^2=S_1 \\ & \therefore 1+36+4-36-a=-16 \\ & \Rightarrow a=5+16=21 \end{aligned} \end{aligned} $

The equation of a circle passing through the intersection of two circles

$ \begin{aligned} & x^2+y^2+2 x+4 y+1=0 \text { and } \\ & x^2+y^2-2 x-4 y-4=0 \text { is } \\ & \left(x^2+y^2+2 x+4 y+1\right) \\ & +\lambda\left(x^2+y^2-2 x-4 y-4\right)=0, \text { where } \lambda \end{aligned} $

$ \begin{aligned} & \text { is a parameter. } \\ & \begin{aligned} \Rightarrow(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) & +(4-4 \lambda) y +1-4 \lambda=0 \\ \Rightarrow x^2+y^2+\frac{2(1-\lambda)}{(1+\lambda)} x & +\frac{4(1-\lambda)}{(1+\lambda)} y +\frac{1-4 \lambda}{1+\lambda}=0 \ldots \text { (i) } \end{aligned} \end{aligned} $

Applying the orthogonal condition,

$ \begin{aligned} 2\left(\frac{1-\lambda}{1+\lambda}\right) \times 0 & +2 \times \frac{2(1-\lambda)}{(1+\lambda)} \times 0 \\ & =\frac{1-4 \lambda}{1+\lambda}-6 \\ \Rightarrow \quad \frac{1-4 \lambda}{1+\lambda} & =6 \\ \Rightarrow \quad 6+6 \lambda & =1-4 \lambda \Rightarrow \lambda=\frac{-1}{2} \end{aligned} $

Substituting $\lambda=\frac{-1}{2}$ in Eq. (i) and simplifying we get

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

$\therefore$ Radius of this circle

$ =\sqrt{3^2+6^2-6}=\sqrt{39} $

Let the $x$-intercepts be $(a, 0)$ and

$ (a+4,0) $

Let the $y$-intercepts be $(0, b)$ and

$ (0, b+2 \sqrt{11}) $

Center of circle $=(a+2, b+\sqrt{11})$

Since, the center is in the fourth quadrant $h>0$ and $k<0$

$ \Rightarrow a+2>0 \text { and } b+\sqrt{11}<0 $

Equation of circle $(x-h)^2+(y-k)^2=r^2$

$ \begin{aligned} & \Rightarrow(1-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2 \\ & \Rightarrow(-a-1)^2+(-b-\sqrt{11})^2=r^2 \end{aligned} $

Substituting $(a, 0)$ gives

$ \begin{aligned} & (a-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2 \\ & \Rightarrow(-2)^2+(-b-\sqrt{11})^2=r^2 \end{aligned} $

substituting ( $a+4,0$ ) gives

$ \begin{aligned} & \Rightarrow (a+4-(a+2))^2+(0-(b+\sqrt{11}))^2=r^2 \\ & \Rightarrow 2^2+(-b-\sqrt{11})^2=r^2 \end{aligned} $

Similarly, $(0, b)$ gives

$ \begin{aligned} (0-(a+2))^2+ & (b-(b+\sqrt{11}))^2=r^2 \\ \text { substituting }(0, b+ & 2 \sqrt{11}) \\ (0-(a+2))^2+(b+ & 2 \sqrt{11} -(b+\sqrt{11}))^2=r^2 \\ (-a-2)^2+(\sqrt{11})^2= & r^2 \end{aligned} $

Solving above equation, we get

$ \begin{aligned} & r^2=20 \\ & r=2 \sqrt{5} \end{aligned} $

The inverse point is $\left(\frac{1}{10}, \frac{-1}{5}\right)$

The original point is $(-1,2)$

The circle equation is

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

Centre of the circle is $(1,-2)$

The original point is $(-1,2)$

$ \begin{aligned} \text { The distance } & =\sqrt{(-1-1)^2+(2-(-2))^2} \\ & =\sqrt{4+16}=\sqrt{20} \end{aligned} $

The centre of the circle is $(1,-2)$

The inverse point is $\left(\frac{1}{10}, \frac{-1}{5}\right)$

$ \begin{aligned} \text { The distance } & =\sqrt{\left(\frac{1}{10}-1\right)^2+\left(\frac{-1}{5}+2\right)^2} \\ & =\frac{9 \sqrt{5}}{10} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & r^2=d_1 d_2=\sqrt{20} \times \frac{9 \sqrt{5}}{10}=9 \\ \Rightarrow & 5-c=9 \Rightarrow c=-4 \end{array} $

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