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} $

$ \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.

$(h,k) = \left( {{{5.5 + 5( - 2) + 14\sqrt 2 } \over {10 + 7\sqrt 2 }},{{35 + 21\sqrt 2 } \over {10 + 7\sqrt 2 }}} )\right.$

$h + k = {{50 + 35\sqrt 2 } \over {10 + 7\sqrt 2 }} = 5$

Equation of AB :

$y - ( - 5) = {{ - 5 + 11} \over {8 - 4}}(x - 8)$

$ \Rightarrow y + 5 = {3 \over 2}(x - 8)$

$ \Rightarrow 2y + 10 = 3x - 24$

$ \Rightarrow 3x - 2y - 34 = 0$ .... (1)

Also AB is cord of contact. And we know equation of cord of contact to a circle is $T = 0$

$ \Rightarrow xh + yk - {3 \over 2}(x + h) + 5(y + k) - 15 = 0$

$ \Rightarrow x\left( {h - {3 \over 2}} \right) + y(k + 5) + \left( { - {3 \over 2}h + 5k - 15} \right) = 0$ .... (2)

Comparing equation (1) and (2), we get

${{h - {3 \over 2}} \over 3} = {{k + 5} \over { - 2}} = {{ - {3 \over 2}h + 5k - 15} \over { - 34}}$

$\therefore$ $ - 2h + 3 = 3k + 15$

$ \Rightarrow 3k + 2h = - 12$ ..... (3)

and

$17(k + 5) = - {3 \over 2}h + 5k - 15$

$ \Rightarrow 12k + {3 \over 2}h - 100$

$ \Rightarrow 3h + 24k = - 200$ ..... (4)

Solving (3) and (4), we get

$h = 8$ and $k = - {{28} \over 3}$

$\therefore$ Point C is $\left( {8, - {{28} \over 3}} \right)$

Now radius of the circle whose centre is at C and tangent is AB is

$ = \left| {{{3(8) - 2\left( { - {{28} \over 3}} \right) - 34} \over {\sqrt {{3^2} + {2^2}} }}} \right|$

$ = \left| {{{26} \over {3\sqrt {13} }}} \right|$

$ = {{2\sqrt {13} } \over 3}$

$\therefore \cos \left(\frac{\theta_{1}}{2}\right)=\frac{r_{i}}{2} \Rightarrow r_{i}=2 \cos \left(\frac{\theta_{i}}{2}\right)$

Given $r_{1}^{2}=r_{2}^{2}+r_{3}^{3}$

$\Rightarrow\left(\cos \left(\frac{\theta_{1}}{2}\right)\right)^{2}=\left(\cos \left(\frac{\theta_{2}}{2}\right)\right)^{2}+\left(\cos \left(\frac{\theta_{3}}{2}\right)\right)^{2}$

$\Rightarrow \frac{3}{4}=\cos ^{2}\left(\frac{\theta_{2}}{2}\right)+\frac{1}{4}$

$\Rightarrow \cos ^{2}\left(\frac{\theta_{2}}{2}\right)=\frac{1}{2}$

$\Rightarrow \frac{\theta_{2}}{2}=\frac{\pi}{4}$

$\Rightarrow \theta_{2}=\frac{\pi}{2}$

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

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

$AO = \sqrt {{{(OC)}^2} - {{(AC)}^2}} $

$ = \sqrt {4 - 1} = \sqrt 3 $

$\sin \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 6}$

Also, $AO = BO$

Area of $\Delta OAB = {1 \over 2}\,.\,OA\,.\,OB\sin 60^\circ $

$ = {1 \over 2} \times \sqrt 3 \,.\,\sqrt 3 \,.\,{{\sqrt 3 } \over 2} = {{3\sqrt 3 } \over 4}$

Let $P(h,k)$ be the orthocentre of $\Delta$ABC

Then

$h = {{\sin t + \cos t + a} \over 3},\,k = {{ - \cos t + \sin t + b} \over 3}$

(Orthocentre coincide with centroid)

$\therefore$ ${(3h - a)^2} + {(3k - b)^2} = 2$

$\therefore$ ${\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}$

$\because$ Orthocentre lies on circle with centre $\left( {1,{1 \over 3}} \right)$

$\therefore$ $a = 3,\,b = 1$

$\therefore$ ${a^2} - {b^2} = 8$

$QR = 2r = 4$

$P = \left( {{1 \over 2} + 2\cos {\pi \over 4}, - 1 + 2\sin {\pi \over 4}} \right)$

$ = \left( {{1 \over 2} + \sqrt 2 , - 1 + \sqrt 2 } \right)$

Area of $\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2 $

$ = 2\sqrt 2 $ sq. units

Equation of C₁

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

Intersection with

$y = 2x$

${x^2} + 4{x^2} - 4x = 0$

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

$ \Rightarrow x = 0,{4 \over 5}$

$y = 0,{8 \over 5}$

$A:\left( {{4 \over 5},{8 \over 5}} \right)$

Tangent of C₂ at $A\left( {{4 \over 5},{8 \over 5}} \right)$

$x + 2y = 4 \Rightarrow P:(4,0),\,Q:(0,2)$

$QA:AP = 1:4$

Circle : ${x^2} + {y^2} - 2gx + 6y - 19c = 0$

It passes through $h(6,1)$

$ \Rightarrow 36 + 1 - 12g + 6 - 19c = 0$

$ = 12g + 19c = 43$ ..... (1)

Line $x - 2cy = 8$ passes through centre

$ \Rightarrow g + 6c = 8$ ...... (2)

From (1) & (2)

$g = 2,\,c = 1$

$C:{x^2} + {y^2} - 4x + 6y - 19 = 0$

x intercept $= 2\sqrt {{g^2} - C} $

$ = 2\sqrt {4 + 19} $

$ = 2\sqrt {23} $

Abscissae of PQ are roots of ${x^2} - 4x - 6 = 0$

Ordinates of PQ are roots of ${y^2} + 2y - 7 = 0$

and PQ is diameter

$\Rightarrow$ Equation of circle is

${x^2} + {y^2} - 4x + 2y - 13 = 0$

But, given ${x^2} + {y^2} + 2ax + 2by + c = 0$

By comparison $a = - 2,b = 1,c = - 13$

$ \Rightarrow a + b - c = - 2 + 1 + 13 = 12$

${c_1}:{x^2} + {y^2} = {r^2}$ ; center = (0, 0) and radius = r

${c_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}$ ; center = (1, 1) and radius = r

${c_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}$ ; center = (2, 1) and radius = r

Distance of $y = mx + c$ line from center (0, 0) is,

$\left| {{{0 + 0 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$ ..... (1)

Distance of $y = mx + c$ line from center (1, 1) is,

$\left| {{{m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$ ..... (2)

Distance of $y = mx + c$ line from center (2, 1) is,

$\left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r$ .... (3)

From (1) and (2), we get

$\left| {{c \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right|$

$ \Rightarrow m - 1 + c = \pm \,c$ ..... (4)

taking positive sign,

$m - 1 + c = c$

$ \Rightarrow m - 1 = 0$

$ \Rightarrow m = 1$

From (2) and (3), we get

$\left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right|$

$ \Rightarrow (m - 1 + c) = \, \pm \,(2m - 1 + c)$ ...... (5)

taking positive sign,

$m - 1 + c = 2m - 1 + c$

$ \Rightarrow m = 0$

By taking positive sign we get two different value of m so it is not acceptable.

From equation (4), taking negative sign,

$m - 1 + c = - c$

$ \Rightarrow m + 2c - 1 = 0$ ..... (6)

From equation (5), taking negative sign

$m - 1 + c = - (2m - 1 + c)$

$ \Rightarrow 3m + 2c - 2 = 0$ ..... (7)

Solving equation (6) and (7), we get

$3m + 1 - m - 2 = 0$

$ \Rightarrow 2m = 1$

$ \Rightarrow m = {1 \over 2}$

$\therefore$ $2c = 1 - {1 \over 2}$

$ \Rightarrow c = {1 \over 4}$

Putting value of $m = {1 \over 2}$ and $c = {1 \over 4}$ in equation (1), we get

$r = \left| {{{{1 \over 4}} \over {\sqrt {1 + {1 \over 4}} }}} \right|$

$ = \left| {{1 \over 4} \times {2 \over {\sqrt 5 }}} \right|$

$ = {1 \over {2\sqrt 5 }}$

$\therefore$ $20({r^2} + c)$

$ = 20\left( {{1 \over {4 \times 5}} + {1 \over 4}} \right)$

$ = 20\left( {{{1 + 5} \over {20}}} \right)$

$ = 6$

Note:

For equation of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$, center is $( - g,\, - f)$ and radius $r = \sqrt {{g^2} + {f^2} - c} $

Given,

equation of circle is

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

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

$ \Rightarrow {x^2} + {y^2} + 2\left( { - {1 \over {\sqrt 2 }}} \right)x + 2\left( { - {1 \over {\sqrt 2 }}} \right) = 0$

$\therefore$ $g = - {1 \over {\sqrt 2 }}$ and $f = - {1 \over {\sqrt 2 }}$

$\therefore$ Center $ = ( - g,\, - f) = \left( {{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$

And Radius $ = r = \sqrt {{{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} + {{\left( { - {1 \over {\sqrt 2 }}} \right)}^2} - 0} $

$ = \sqrt {{1 \over 2} + {1 \over 2}} = \sqrt 1 = 1$

As AB and AC makes an angle 90$^\circ$ then line BC passes through the center of circle and BC is the diameter of the circle.

$\therefore$ Length of BC = 2r = 2 $\times$ 1 = 2

$\therefore$ AC² = BC² $-$ AB²

= 2² $-$ ($\sqrt2$)²

= 2

$\Rightarrow$ AC = $\sqrt2$

$\therefore$ Area of right angle triangle ABC

= ${1 \over 2}$ $\times$ AC $\times$ AB

= ${1 \over 2}$ $\times$ $\sqrt2$ $\sqrt2$

= 1 square unit.

Equation of tangent at point M is

$T = 0$

$ \Rightarrow x{x_1} + y{y_1} = 2$

$ \Rightarrow - x + y = 2$

$ \Rightarrow y = x + 2$

Putting this value to equation of circle C₂,

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

$ \Rightarrow {(x - 3)^2} + {x^2} = 5$

$ \Rightarrow {x^2} - 6x + 9 + {x^2} = 5$

$ \Rightarrow 2{x^2} - 6x + 4 = 0$

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

$ \Rightarrow (x - 2)(x - 1) = 0$

$ \Rightarrow x = 1,2$

when $x = 1,$ $y = 3$

and when $x = 2,$ $y = 4$

$\therefore$ Point $A(1,3)$ and $B(2,4)$

Now, equation of tangent at $A(1,3)$ on circle ${(x - 3)^2} + {(y - 2)^2} = 5$ or ${x^2} + {y^2} - 6x - 4y + 8 = 0$ is

$T = 0$

$x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + C = 0$

$ \Rightarrow x + 3y - 3(x + 1) - 2(y + 3) + 8 = 0$

$ \Rightarrow x + 3y - 3x - 3 - 2y - 6 + 8 = 0$

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

$ \Rightarrow 2x - y + 1 = 0$ ....... (1)

Similarly tangent at $B(2,4)$ is

$2x + 4y - 3(x + 2) - 2(y + 4) + 8 = 0$

$ \Rightarrow 2x + 4y - 3x - 6 - 2y - 8 + 8 = 0$

$ \Rightarrow - x + 2y - 6 = 0$

$ \Rightarrow x - 2y + 6 = 0$ ...... (2)

Solving equation (1) and (2), we get

$x - 2(2x + 1) + 6 = 0$

$ \Rightarrow x - 4x - 2 + 6 = 0$

$ \Rightarrow - 3x + 4 = 0$

$ \Rightarrow x = {4 \over 3}$

$\therefore$ $y = 2 \times {4 \over 3} + 1 = {{11} \over 3}$

$\therefore$ Point $N = \left( {{4 \over 3},{{11} \over 3}} \right)$

Now area of the triangle ANB

$ = {1 \over 2}\left| {\matrix{ {{x_1}} & {{y_1}} & 1 \cr {{x_2}} & {{y_2}} & 1 \cr {{x_3}} & {{y_3}} & 1 \cr } } \right|$

$ = {1 \over 2}\left| {\matrix{ 1 & 3 & 1 \cr 2 & 4 & 1 \cr {{4 \over 3}} & {{{11} \over 3}} & 1 \cr } } \right|$

$ = {1 \over 2}\left[ {1\left( {4 - {{11} \over 3}} \right) - 3\left( {2 - {4 \over 3}} \right) + 1\left( {{{22} \over 3} - {{16} \over 3}} \right)} \right]$

$ = {1 \over 2}\left[ {{1 \over 3} - {6 \over 3} + {6 \over 3}} \right]$

$ = {1 \over 6}$

$\tan 2\theta = 2 \Rightarrow {{2\tan \theta } \over {1 - {{\tan }^2}\theta }} = 2$

$\tan \theta = {{\sqrt 5 - 1} \over 2}$ (as $\theta$ is acute)

Area $ = {1 \over 2}{L^2}\sin 2\theta = {1 \over 2}.\,{5 \over {{{\tan }^2}\theta }}.\,2\sin \theta \cos \theta $

$ = {{5\sin \theta \cos \theta } \over {{{\sin }^2}\theta }}.\,{\cos ^2}\theta $

$ = 5\cot \theta .\,{\cos ^2}\theta $

$ = 5.\,{2 \over {\sqrt 5 - 1}}.\,{1 \over {1 + {{\left( {{{\sqrt 5 - 1} \over 2}} \right)}^2}}}$

$ = {{10} \over {\sqrt 5 - 1}}.\,{4 \over {4 + 6 - 2\sqrt 5 }}$

$ = {{40} \over {2\sqrt 5 {{(\sqrt 5 - 1)}^2}}} = {{4\sqrt 5 } \over {6 - 2\sqrt 5 }}$

$ = {{4\sqrt 5 (6 + 2\sqrt 5 )} \over {16}}$

$ = {{\sqrt 5 (3 + \sqrt 5 )} \over 2}$