Circle

The circle is $x^2+y^2=2^2$ (radius $r=2$ ).

• Intersection with $\mathbf{x}$-axis: Setting $y=0$ gives $x^2=4 \Rightarrow x= \pm 2$.

• Since $a>0, A=(2,0)$.

• Therefore, $B=(-2,0)$.

• Points $P$ and $Q$ :

• $P=(2 \cos \alpha, 2 \sin \alpha)$

• $Q=(2 \cos \beta, 2 \sin \beta)$

Let point of intersection of $A Q$ .& $B P$ be $R(h, k)$.

So, we need to find locus of points of intersection of $A Q \& B P$.

Slope of $B P=$ slope of $B R$

$ \begin{gathered} \frac{k-0}{h-(-2)}=\frac{2 \sin \alpha-0}{2 \cos \alpha+2} \\ \frac{k}{h+2}=\frac{\sin \alpha}{\cos \alpha+1} \Rightarrow \frac{k}{h+2}=\frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^2 \frac{\alpha}{2}} \\ \frac{k}{h+2}=\tan \frac{\alpha}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{gathered} $

Similarly, slope of $Q A=$ slope of $Q R$

$ \begin{gathered} \frac{2 \sin \beta-0}{2 \cos \beta-2}=\frac{k-0}{h-2} \\ \frac{\sin \beta}{\cos \beta-1}=\frac{k}{h-2} \Rightarrow \frac{2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}}{-2 \sin ^2 \frac{\beta}{2}}=\frac{k}{h-2} \\ -\cot \frac{\beta}{2}=\frac{k}{h-2} \\ \cot \frac{\beta}{2}=\frac{k}{2-h} \Rightarrow \tan \frac{\beta}{2}=\frac{2-h}{k} \end{gathered} $

Given,

$ \begin{gathered} \alpha-\beta=\frac{\pi}{2} \Rightarrow \frac{\alpha}{2}-\frac{\beta}{2}=\frac{\pi}{4} \\ \tan \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=\tan \frac{\pi}{4} \\ \frac{\tan \frac{\alpha}{2}-\tan \frac{\beta}{2}}{1+\tan \frac{\alpha}{2} \tan \frac{\beta}{2}}=1 \end{gathered} $

Substitute

$ \tan \frac{\alpha}{2}=\frac{k}{h+2}, \quad \tan \frac{\beta}{2}=\frac{2-h}{k} $

$ \begin{gathered} \frac{\frac{k}{h+2}-\frac{2-h}{k}}{1+\left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right)}=1 \\ \frac{k}{h+2}+\frac{h-2}{k}=1+\left(\frac{k}{h+2}\right)\left(\frac{2-h}{k}\right) \\ \frac{k^2+h^2-4}{(h+2) k}=\frac{4 k}{(h+2) k} \\ \frac{k^2+h^2-4}{(h+2) k}=\frac{4 k}{(h+2) k} \\ k^2+h^2-4=4 k \\ k^2+h^2-4 k-4=0 \end{gathered} $

Replace $(h, k) \rightarrow(x, y)$

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

To find the value of $a-b$,

Determine the equation of circle $C_1$

The circle $C_1$ has a diameter of 10 , so its radius is $R=5$. It passes through the origin ( 0,0 ) and is contained in the closed half-plane $x \geq 0$.

Let the center of $C_1$ be $(h, k)$. Since it passes through $(0,0)$, we have $h^2+k^2=R^2=25$.

For the circle to lie entirely in the region $x \geq 0$ while passing through $(0,0)$, the distance from the center to the $y$-axis must be equal to the radius, and the center must have a non-negative $x$-coordinate.

Thus, $h=R=5$.

Substituting $h=5$ into $h^2+k^2=25$ gives $25+k^2=25$, which implies $k=0$.

The equation of $C_1$ is:

$ (x-5)^2+y^2=25 $

Find the center of circle $C_2$

The line $y=x$ is a chord of $C_1$. We find the intersection points of $y=x$ and $C_1$ :

$(x-5)^2+x^2 =25 $

$\Rightarrow $ $ x^2-10 x+25+x^2 =25 $

$\Rightarrow $$ 2 x^2-10 x =0 $

$\Rightarrow $ $ 2 x(x-5)=0 $

The intersection points are $(0,0)$ and $(5,5)$. These are the endpoints of the diameter of circle $C_2$.

The center $M$ of $C_2$ is the midpoint of the chord:

$ M=\left(\frac{0+5}{2}, \frac{0+5}{2}\right)=(2.5,2.5) $

Determine the equation of the chord of $C_2$

We are looking for the equation of a chord of $C_2$ passing through $P(2,3)$ that is farthest from the center $M(2.5,2.5)$.

In any circle, the chord passing through a point $P$ that is farthest from the center is the one perpendicular to the radius (or segment) $M P$.

The slope of the segment $M P$ is:

$ m_{M P}=\frac{3-2.5}{2-2.5}=\frac{0.5}{-0.5}=-1 $

The slope $m$ of the required chord is the negative reciprocal of $m_{M P}$ :

$ m=-\frac{1}{-1}=1 $

Using the point-slope form for the line through $P(2,3)$ with slope $m=1$ :

$ y-3 =1(x-2) $

$\Rightarrow $ $ y-3 =x-2 $ $\Rightarrow $ $ x-y+1 =0 $

Calculate $a-b$

The equation is given in the form $x+a y+b=0$. Comparing this to $x-y+1=0$, we identify:

$ \begin{aligned} & a=-1 \\ & b=1 \end{aligned} $

Finally, we calculate the required value:

$ a-b=-1-1=-2 $

Given a circle passes through $O(0,0), A(-\sqrt{3} a, 0)$, and $B(0,-\sqrt{2} b)$. Since $A$ lies on the x -axis and $B$ lies on the y -axis, $\angle A O B=90^{\circ}$. Thus $A B$ is the diameter of the circle.

Diameter $d=2 \times$ radius $=2(4)=8$.

Using distance formula for $A B$ :

$ \begin{gathered} \sqrt{(-\sqrt{3} a-0)^2+(0-(-\sqrt{2} b))^2}=8 \\ 3 a^2+2 b^2=64 \quad------(1) \end{gathered} $

Let the centroid of $\triangle O A B$ be $G(h, k)$.

$ \begin{gathered} h=\frac{0-\sqrt{3} a+0}{3} \Longrightarrow a=\frac{-3 h}{\sqrt{3}}=-\sqrt{3} h \\ k=\frac{0+0-\sqrt{2} b}{3} \Longrightarrow b=\frac{-3 k}{\sqrt{2}} \end{gathered} $

Substitute $a$ and $b$ into (Eq. 1):

$ \begin{gathered} 3(-\sqrt{3} h)^2+2\left(\frac{-3 k}{\sqrt{2}}\right)^2=64 \\ 3\left(3 h^2\right)+2\left(\frac{9 k^2}{2}\right)=64 \\ 9 h^2+9 k^2=64 \\ h^2+k^2=\frac{64}{9} \end{gathered} $

The locus is a circle $x^2+y^2=\left(\frac{8}{3}\right)^2$.

Radius $=\frac{8}{3} $

$ \begin{aligned} & \left|r_1-r_2\right| < c_1 c_2 < r_1+r_2 \\ & r-3 < \sqrt{9+25} < r+3 \\ & r \in(\sqrt{34}-3, \sqrt{34}+3) \\ & \alpha \beta=25 \end{aligned} $

$ \begin{aligned} & x^2+y^2-14 x-6 y-3=0 \\ & \tan 60^{\circ}=\frac{4}{\sqrt{h^2+k^2-4 h-6 k-3}} \end{aligned} $

squaring both side

$ 3\left(h^2+k^2-4 h-6 k-3\right)=16 $

To get locus replace $h, k$ by $x$ and $y$;

$ \begin{aligned} & 3 x^2+3 y^2-12 x-18 y-9-16=0 \\ & 3 x^2+3 y^2-12 x-18 y-25=0 \end{aligned} $

Since the circle touches the $x$-axis at a point which is $3$ units from the origin, the point of contact is $(3,0)$.

The centre lies in the first quadrant and the radius is perpendicular to the tangent at the point of contact. So, the centre must be of the form $(3,r)$, where $r$ is the radius.

The circle has an intercept of length $6\sqrt{3}$ on the $y$-axis. This means the chord cut by the line $x=0$ has length $6\sqrt{3}$.

For a circle of radius $r$, if the distance of a chord from the centre is $d$, then chord length is $2\sqrt{r^2-d^2}$.

Here, the distance from centre $(3,r)$ to the $y$-axis $(x=0)$ is $d=3$. So the intercept on the $y$-axis is

$2\sqrt{r^2-3^2}=6\sqrt{3}$

So,

$\sqrt{r^2-9}=3\sqrt{3}\ \Rightarrow\ r^2-9=27\ \Rightarrow\ r^2=36\ \Rightarrow\ r=6$

Hence, centre of the circle is $(3,6)$ and radius is $6$ units.

Now we need the length of the chord cut by the line $x-y=3$.

First find the perpendicular distance $p$ from the centre $(3,6)$ to the line $x-y-3=0$:

$p=\frac{|1\cdot 3+(-1)\cdot 6-3|}{\sqrt{1^2+(-1)^2}}=\frac{|3-6-3|}{\sqrt{2}}=\frac{6}{\sqrt{2}}$

Now chord length on this line is

$P Q=2 \sqrt{6^2-p^2} \quad \therefore p=\frac{6}{\sqrt{2}}$

$\begin{aligned} & \mathrm{p}=\frac{|3-6-3|}{\sqrt{2}} \\ & \mathrm{PQ}=2 \sqrt{36-\frac{36}{2}} \\ & =2 \times \frac{6}{\sqrt{2}}=6 \sqrt{2} \end{aligned}$

Vertex of the parabola

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

is found by completing the square:

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

So, the vertex is

$ P=(3,3) $

Now the circle is

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

Complete squares:

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

Hence, the centre of the circle is

$ C=(1,2) $

and radius is

$ r=\sqrt{2} $

Now,

$ PC=\sqrt{(3-1)^2+(3-2)^2}=\sqrt{4+1}=\sqrt{5} $

Since the line through $P$ cuts the circle at $R$ and $S$, the line must intersect the circle in a chord $RS$.

Let the perpendicular distance from the centre $C$ to this line be $d$.

Then chord length

$ RS=2\sqrt{r^2-d^2}=2\sqrt{2-d^2} $

Also, since $P$ lies on the line outside the circle, and $R,S$ are the intersection points on opposite sides along the line, we have

$ PR+PS=RS $

only if $P$ lies between them, which is not possible here because $P$ is outside the circle.

So let us use power of a point.

If $PR$ and $PS$ are the two secant segments from $P$, then

$ PR \cdot PS = PC^2-r^2=5-2=3 $

Let the line make an angle such that the foot from $C$ to the line gives distance $d$.

If $M$ is the midpoint of chord $RS$, then

$ CM=d,\qquad PM=\sqrt{PC^2-d^2}=\sqrt{5-d^2} $

because in right triangle with hypotenuse $PC$.

Also

$ MR=MS=\sqrt{r^2-d^2}=\sqrt{2-d^2} $

Since $P$ is outside the circle, along the secant line,

$ PR=PM-MR,\qquad PS=PM+MR $

Therefore,

$ PR+PS = 2PM = 2\sqrt{5-d^2} $

So

$ (PR+PS)^2=4(5-d^2) $

To maximize this, we must minimize $d^2$.

The minimum possible value of $d$ is $0$, when the line passes through the centre $C$.

Hence maximum value is

$ (PR+PS)^2_{\max}=4\times 5=20 $

Therefore, the required maximum value is

$ \boxed{20} $

So the correct option is B.

First, find the vertex of the parabola.

The parabola is

$ x^2+6x+y+13=0 $

Rewrite it by completing the square:

$ x^2+6x+y+13=0 $

$ x^2+6x+9+y+13-9=0 $

$ (x+3)^2+y+4=0 $

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

So, the vertex of the parabola is

$ (-3,-4) $

Now, find the centre and radius of the circle.

Given circle:

$ x^2+y^2-6x-8y+21=0 $

Complete the squares:

$ x^2-6x+y^2-8y+21=0 $

$ (x^2-6x+9)+(y^2-8y+16)+21-9-16=0 $

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

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

Hence, centre of the circle is

$ (3,4) $

and radius is

$ r=2 $

Now we need the maximum distance of a point $P$ on this circle from the vertex $(-3,-4)$.

Let $V=(-3,-4)$ and centre of circle be $C=(3,4)$.

First find $VC$:

$ VC=\sqrt{(3+3)^2+(4+4)^2} =\sqrt{6^2+8^2} =\sqrt{36+64} =\sqrt{100} =10 $

For a point moving on a circle, the maximum distance from an external point is

$ VC+r $

So,

$ \text{Maximum distance}=10+2=12 $

Therefore, the correct answer is

$ \boxed{12} $

So, Option C is correct.

The given circle is

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

and the point from which the two chords are drawn is

$ P(1,2) $

We are told that these chords are bisected by the $y$-axis.

Let the other end of one such chord be $R(x,y)$.

Since the chord joining $P(1,2)$ and $R(x,y)$ is bisected by the $y$-axis, its midpoint must lie on the $y$-axis.

So the $x$-coordinate of the midpoint is $0$.

Midpoint of $P(1,2)$ and $R(x,y)$ is

$ \left(\frac{1+x}{2}, \frac{2+y}{2}\right) $

Since this lies on the $y$-axis,

$ \frac{1+x}{2}=0 $

So,

$ x=-1 $

Thus, the other end of each such chord must lie on the line

$ x=-1 $

Now find where this line cuts the circle.

Substitute $x=-1$ in

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

We get

$ (-1)^2+y^2+(-1)-3y=0 $

$ 1+y^2-1-3y=0 $

$ y^2-3y=0 $

$ y(y-3)=0 $

So,

$ y=0 \quad \text{or} \quad y=3 $

Hence the two points are

$ R(-1,0), \quad S(-1,3) $

Now midpoint of $RS$ is

$ (\alpha,\beta)=\left(\frac{-1+(-1)}{2},\frac{0+3}{2}\right) $

$ (\alpha,\beta)=\left(-1,\frac{3}{2}\right) $

Therefore,

$ \alpha+\beta=-1+\frac{3}{2}=\frac{1}{2} $

So,

$ 6(\alpha+\beta)=6\cdot \frac{1}{2}=3 $

Hence the correct answer is

$ \boxed{3} $

So, the correct option is Option B.

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

Solving :

$ \begin{aligned} & 2(x-2 y+3)=0 \\ & 2 x+y-4=0 \\ & -5 y+10=0 \\ & y=2 \end{aligned} $

$ x=1 $

Centre $(1,2)$

$ r=\sqrt{5} $

So circle is $(x-1)^2+(y-2)^2=5$

Chord $\mathrm{x}-\mathrm{y}+2=0$

$ \begin{aligned} & \mathrm{p}=\frac{1}{\sqrt{2}} \\ & \ell=\sqrt{5-\frac{1}{2}}=\frac{3}{\sqrt{2}} \\ & \mathrm{AB}=\frac{6}{\sqrt{2}} \quad \therefore \mathrm{AB}^2=18 \end{aligned} $

$ \begin{aligned} & \sqrt{16+36}=r+3 \\ & \Rightarrow r=\sqrt{52}-3 \\ & \Rightarrow \frac{3-3 r}{-r+3}=\alpha, \beta=\frac{9-3 r}{3+r} \\ & \Rightarrow \frac{(9-3 r-3+3 r)^2}{(r+3)^2} \\ & \Rightarrow \frac{36}{12}=\frac{9}{13}=\frac{m}{n} \Rightarrow m+n=22 \end{aligned}$

$\begin{aligned} 2 r & =\sqrt{52} \\ 4 r^2 & =52 \end{aligned}$

$\begin{aligned} &\begin{aligned} & 2 r^2=26 \\ & r^2=13 \end{aligned}\\ &\text { Eq }{ }^{\mathrm{n}} \text { of circle is }\\ &\begin{aligned} & x(x-4)+y(y-6)=0 \\ & k(k-4)+3 k(3 k-6)=0 \\ & k^2-4 k+9 k^2-18 k=0 \\ & 10 k^2=22 k \\ & 10 k=22 \\ & \therefore \quad 10 k+r^2=35 \end{aligned} \end{aligned}$

$\begin{aligned} &\begin{aligned} & \mathrm{M}_{A B}=0 \Rightarrow \mathrm{OM} \text { is vertical } \\ & \Rightarrow \alpha=2 \\ & \therefore \text { Centre }(0) \equiv(2,-4) \\ & \quad r=O A=\sqrt{(2-4)^2+(2+4)^2}=\sqrt{40} \end{aligned}\\ &\text { mid point of chord is } \mathrm{N} \equiv(1,2) \quad \therefore \mathrm{ON}=\sqrt{37}\\ &\begin{aligned} \therefore \text { length of chord } & =2 \sqrt{\mathrm{r}^2-(\mathrm{ON})^2} \\ & =2 \sqrt{40-37}=2 \sqrt{3} \end{aligned} \end{aligned}$

By solving $\mathrm{x}=\mathrm{y}$ with circle

We get

$\begin{aligned} & \mathrm{C}(\sqrt{2}, \sqrt{2}) \\ & \mathrm{D}(-\sqrt{2},-\sqrt{2}) \end{aligned}$

By solving $\mathrm{x}+\mathrm{y}=1$ with circle $x^2+y^2=4$ we set

$\mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right)$

& $\mathrm{B}\left(\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}\right)$

$\therefore$ Area of Quadrilateral ACBD

$=2 \times$ Area of $\triangle \mathrm{BCD}$

$\begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1 \end{array}\right| \\ & =2 \sqrt{14} \end{aligned}$

By pytogorus $\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2$

$r=\sqrt{\frac{4 a^2+b^2}{4}}$

Equation of circle is $(x-\alpha)^2+(y-\beta)^2=r^2$

$\begin{aligned} & x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0 \\ & \text { comparision } x^2+y^2-\alpha x+\beta y+r=0 \\ & -\alpha=-2 a, \beta=-2 p, r=a^2 \\ & \Rightarrow 2 a=\alpha, 4 a^2+b^2=4 p^2 \\ & \alpha^2+b^2=4 p^2 \\ & \alpha^2+b^2=\beta^2 \end{aligned}$

So, $\left(2 a, b^2\right)=\left(\alpha, \beta^2-4 r\right)$

Centre $(1,-2), r=3$

Reflection of $(1,-2)$ about $2 x-3 y+5=0$

$\begin{aligned} & \frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2 \\ & x=-3, y=4 \end{aligned}$

Equation of circle ' $C$ '

$C:(x+3)^2+(y-4)^2=9$

A.T.Q.

$\begin{aligned} & \ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r} \\ & \mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r} \\ & \theta=\frac{\pi}{3} \\ & (\alpha+6)^2+(\beta-4)^2=27 \\ & \frac{(\alpha+3)^2 \pm(\beta-4)^2=9}{(\alpha+6)^2-(\alpha+3)^2=18} \\ & \Rightarrow 6 \alpha=-9 \\ & \Rightarrow \alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right) \\ & \therefore \beta-\sqrt{3} \alpha \\ & \left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2} \\ & =4 \end{aligned}$

$ S_1:(x+2)^2+(y-2)^2=2^2 $

$ \mathrm{S}_2:(\mathrm{x}-2)^2+(\mathrm{y}-5)^2=\mathrm{r}^2 $

Both circle intersect at two points

$ \begin{aligned} & \therefore\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{c}_{\mathrm{c}} \mathrm{c}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & |\mathrm{r}-2|<5<2+\mathrm{r} \\ & \Rightarrow 3<\mathrm{r}<7 \\ & \mathrm{r} \in(3,7) \\ & \alpha=3, \beta=7 \\ & 3 \beta-2 \alpha=15 \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)

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

We are given two circles:

(1) $x^2+y^2-18 x-15 y+131=0$

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

First, let's find the centers and radii of the circles.

For circle (1):

Completing the square for the equation:

$(x^2-18x+{81})+(y^2-15y+\frac{225}{4})=-131+{81}+\frac{225}{4}$

$(x-9)^2+(y-\frac{15}{2})^2=\frac{25}{4}$

Center 1: $C_1(9, \frac{15}{2})$

Radius 1: $r_1 = \sqrt{\frac{25}{4}}=\frac{5}{2}$

For circle (2):

Completing the square for the equation:

$(x^2-6x+9)+(y^2-6y+9)=7+9+9$

$(x-3)^2+(y-3)^2=25$

Center 2: $C_2(3, 3)$

Radius 2: $r_2 = 5$

Now, let's find the distance between the centers :

$d = \sqrt{(9-3)^2 + (\frac{15}{2}-3)^2} = \sqrt{6^2 + \frac{9}{2}^2} = \sqrt{36 + \frac{81}{4}} = \frac{15}{2}$

Next, let's analyze the relative positions of the circles using the distance between centers and the sum and difference of the radii :

1. If $d > r_1 + r_2$, the circles are separate, and there are 4 common tangents.
2. If $d = r_1 + r_2$, the circles are externally tangent, and there are 3 common tangents.
3. If $0 < d < |r_1 - r_2|$, one circle is inside the other, and there are no common tangents.
4. If $d = |r_1 - r_2|$, the circles are internally tangent, and there is 1 common tangent.
5. If $d < |r_1 - r_2|$, one circle is completely inside the other, and there are no common tangents.

In this case :

$d = \frac{15}{2}$

$r_1 = \frac{5}{2}$

$r_2 = 5$

Now, let's check the conditions :

$r_1 + r_2 = \frac{5}{2} + 5$ = $\frac{15}{2}$

Since $d = \frac{15}{2} = r_1 + r_2$, the circles touch each other externally, and there are 3 common tangents.