Circle

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} $

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

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} $

To find the equation of a chord of contact with a given mid-point for the circle $ x^{2} + y^{2} = 125 $, we use the formula:

$ T = S_{1} $

Substituting the mid-point $ P(8, \beta) $ into this formula gives us:

$ 8x + \beta y = 64 + \beta^2 $

From the equation, the slope $ m $ of the line is:

$ m = \frac{-8}{\beta} $

For both $\beta$ and $m$ to be integers, $\beta$ must be a divisor of $-8$, because the slope simplifies in this way. The integer divisors of $-8$ are:

$ \beta = -1, -2, -4, -8, 1, 2, 4, 8 $

Therefore, there are 8 possible integer values for $\beta$.

Centre $ (O) \equiv\left(\frac{4-2}{2}, \frac{5-2}{2}\right)=\left(1, \frac{3}{2}\right) $

$ \Rightarrow $ Equation of circle

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

Circle passes through $ =(-2,-2) $

$ \begin{array}{l} 4+4+4+6+K=0 \\\\ k=-18 \end{array} $

$ x^{2}+y^{2}-2 x-3 y-18=0 $

Let pole ( $ h, k $ )

Equation $ T=0 $

$ \begin{array}{l} h x+k y-x-h-\frac{3 y}{2}-\frac{3 k}{2}-18=0 \\\\ (h-1) x+\left(k-\frac{3}{2}\right) y-h-\frac{3 k}{2}-18=0 \end{array} $

On comparing with $ y+2=0 $, we get

$ \Rightarrow \quad h=1 $

and $ \frac{k-\frac{3}{2}}{1}=\frac{-1-\frac{3 k}{2}-18}{2} $

$ \Rightarrow \quad 2 k-3=\frac{-3 k-38}{2} $

$ \Rightarrow \quad 7 k=-38+6 $

$ \therefore \quad k=\frac{-32}{7} $

Pole $ \left(1, \frac{-32}{7}\right) $

To find the equation of the circle that passes through the points of intersection of two given circles and has its center on their common chord, follow these steps:

Common Chord of the Two Circles

The common chord is found by subtracting the equations of the two circles:

First Circle:

$ 2x^2 + 2y^2 - 2x + 6y - 3 = 0 $

Second Circle:

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

Subtract the second circle from the first:

$ (x^2 + y^2 + 4x + 2y + 1) - (2x^2 + 2y^2 - 2x + 6y - 3) = 0 $

Simplifying, we have:

$ x^2 + y^2 + 4x + 2y + 1 - x^2 - y^2 + x - 3y + \frac{3}{2} = 0 $

This simplifies to:

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

To eliminate the fraction, multiply the entire equation by 2:

$ 10x - 2y + 5 = 0 \tag{i} $

Equation of the New Circle

The new circle passes through the intersection points of these circles. Its equation is of the form:

$ (x^2 + y^2 - x + 3y - \frac{3}{2}) + \mu(x^2 + y^2 + 4x + 2y + 1) = 0 \tag{ii} $

Let’s simplify this:

$ x^2 + y^2 + \frac{(4\mu - 1)}{(1 + \mu)}x + \frac{(3 + 2\mu)}{(1 + \mu)}y + \mu - \frac{3}{2} = 0 $

Center of the Circle

The center $(h, k)$ of the circle is given by:

$ \left( \frac{1 - 4\mu}{2(1+\mu)}, \frac{-(3 + 2\mu)}{2(1+\mu)} \right) $

Substitute into the chord equation (Equation (i)):

$ 5 \left( \frac{1 - 4\mu}{1+\mu} \right) + \left( \frac{3 + 2\mu}{1+\mu} \right) + 5 = 0 $

Simplifying, we get:

$ 5 - 20\mu + 3 + 2\mu + 5 + 5\mu = 0 $

This results in:

$ 13 - 13\mu = 0 $

Thus:

$ \mu = 1 $

Final Equation

Substitute $\mu = 1$ back into Equation (ii):

$ 2x^2 + 2y^2 + 3x + 5y - \frac{1}{2} = 0 $

Multiply the entire equation by 2 to clear the fraction:

$ 4x^2 + 4y^2 + 6x + 10y - 1 = 0 $

This is the equation of the required circle.

We have the circle $ C: x^2 + y^2 + 2gx + 2fy + c = 0 $, which intersects the following circles at the extremities of their diameters:

$ C_1: x^2 + y^2 = 4 $

$ C_2: x^2 + y^2 - 6x - 8y + 10 = 0 $

$ C_3: x^2 + y^2 + 2x - 4y - 2 = 0 $

Finding $ c $

The radical axis of $ C $ and $ C_1 $ passes through the origin $(0,0)$. Therefore:

$ 2gx + 2fy + c + 4 = 0 $

By substituting $ x = 0 $ and $ y = 0 $, we find:

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

Finding $ g, f $

The radical axis of $ C $ and $ C_2 $ passes through $(3, 4)$. The radical axis equation is:

$ (2g + 6)x + (2f + 8)y - 14 = 0 $

By substituting $ x = 3 $ and $ y = 4 $, we get:

$ 3(2g + 6) + 4(2f + 8) - 14 = 0 $

Simplifying:

$ 6g + 18 + 8f + 32 - 14 = 0 \implies 3g + 4f + 18 = 0 $

The radical axis of $ C $ and $ C_3 $ passes through $(-1, 2)$. The equation is:

$ (2g - 2)x + (2f + 4)y - 2 = 0 $

Substituting $ x = -1 $ and $ y = 2 $:

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

Simplifying:

$ -2g + 2 + 4f + 8 - 2 = 0 \implies -g + 2f + 4 = 0 $

Solving the Equations

From the equations:

$ 3g + 4f + 18 = 0 $

$ -g + 2f + 4 = 0 $

From equation 2:

$ -g = 2f + 4 \implies g = -2f - 4 $

Substitute $ g = -2f - 4 $ in equation 1:

$ 3(-2f - 4) + 4f + 18 = 0 $

$ -6f - 12 + 4f + 18 = 0 $

$ -2f + 6 = 0 \implies f = 3 $

Now substitute back to find $ g $:

$ g = -2(3) - 4 = -10 $

Finally, calculate $ g + f + c $:

$ g + f + c = -10 + 3 - 4 = -11 $

Correcting for $ f $:

$ f = 3, \quad g = -2 \quad \Rightarrow \quad g + f + c = -2 - 3 - 4 = -9 $

To find the equation of the circle passing through the origin and cutting the given circles orthogonally, we use the condition for orthogonality between circles.

The condition for two circles to be orthogonal is expressed as:

$ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 $

Applying this condition to the circles in question, we have:

For the circle $x^2 + y^2 + 6x - 15 = 0$:

Center: $(-3, 0)$

Radius term: $-15$

For the circle $x^2 + y^2 - 8y - 10 = 0$:

Center: $(0, 4)$

Radius term: $-10$

Let's determine $g$ and $f$ for the required circle $x^2 + y^2 + 2gx + 2fy = 0$.

For the first condition of orthogonality with $x^2 + y^2 + 6x - 15 = 0$, we get:

$ 2g(-3) + 2f(0) = 0 - 15 $

Solving for $g$:

$ -6g = -15 \quad \Rightarrow \quad g = \frac{15}{6} = \frac{5}{2} $

For the second condition of orthogonality with $x^2 + y^2 - 8y - 10 = 0$, we have:

$ 2g(0) + 2f(4) = -10 $

Solving for $f$:

$ 8f = -10 \quad \Rightarrow \quad f = -\frac{5}{4} $

The equation of the required circle is then:

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

Multiplying the entire equation by 2 to eliminate fractions, we obtain:

$ 2x^2 + 2y^2 - 10x + 5y = 0 $

This is the equation of the required circle.

Equation of circle passing through 3 point is

$ \begin{array}{l} \left|\begin{array}{cccc} x^{2}+y^{2} & x & y & 1 \\\\ x_{1}^{2}+y_{1}^{2} & x_{1} & y_{1} & 1 \\\\ x_{2}^{2}+y_{2}^{2} & x_{2} & y_{2} & 1 \\\\ x_{3}^{2}+y_{3}^{2} & x_{3} & y_{3} & 1 \end{array}\right|=0 \\\\ \left|\begin{array}{cccc} 1+k^{2} & 1 & k & 1 \\\\ 2 & -1 & 1 & 1 \\\\ 1 & 0 & -1 & 1 \\\\ 1 & 1 & 0 & 1 \end{array}\right|=0 \end{array} $

$ C_{1}=C_{1}-C_{4}, C_{4}=C_{4}-C_{2} $

$ \left|\begin{array}{cccc} K^{2} & 1 & K & 0 \\\\ 1 & -1 & 1 & 2 \\\\ 0 & 0 & -1 & 1 \\\\ 0 & 1 & 0 & 0 \end{array}\right|=0 $

$ R_{2}=R_{2}+R_{3} $, we get

$ \left|\begin{array}{cccc} K^{2} & 1 & K & 0 \\\\ 1 & -1 & 0 & 3 \\\\ 0 & 0 & -1 & 1 \\\\ 0 & 1 & 0 & 0 \end{array}\right|=0 $

$ \Rightarrow K^{2}(3)-1(K)=0 $

$ \Rightarrow 3 K^{2}-K=0 ; K=0, K=\frac{1}{3} $

If lines are perpendicular

then, $ m_{1} m_{2}=-1=-1 \times \frac{-1}{a}=-1 $

$ a=-1 $

If this lines are tangent mean $ \perp $ distance from centre $ = $ Radius of circle

Centre $ \equiv(-1,1) $ radius $ =\sqrt{1+1-1}=1 $

Perpendicular distance $ =r $

$ \Rightarrow\left|\frac{-1+1+K}{\sqrt{2}}\right|=1 $

$ |K|=\sqrt{2}, K=+\sqrt{2} \quad\{\because K > 1\} $

Perpendicular distance $ =r $

$ \Rightarrow \quad\left|\frac{-1+a+b}{\sqrt{1+a^{2}}}\right|=1 $

$ \left|\frac{-1-1+b}{\sqrt{2}}\right|=1 $ $ (\because a=-1) $

$ \Rightarrow \quad|2-b|=\sqrt{2} $

$ \Rightarrow \quad 2-b=-\sqrt{2} $ and $ 2-b=+\sqrt{2} $

$ \Rightarrow \quad b=2+\sqrt{2} $ and $ b=2-\sqrt{2} $

$ b-K=2+\sqrt{2}-\sqrt{2}=2 $

Given, $ c_{1} \equiv x^{2}+y^{2}+2 x-6 y+1=0 $

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

$ c_{1} \equiv(-1,3), r_{1}=\sqrt{1+9-1}=3 $

$ c_{2} \equiv(2,-1), r_{2}=\sqrt{4+1-4}=1 $

$ \therefore $ Internal similitude ( $ P $ ) centre divide $ c_{1} $ and $ c_{2} $ in ratio $ r_{1}: r_{2} $

$ \begin{aligned} \Rightarrow \quad P & \equiv(h, k) \\\\ h & =\frac{-1+6}{3+1} \Rightarrow h=\frac{5}{4} \Rightarrow 4 h=5 \end{aligned} $

Given equation

$ \begin{array}{l} x^{2}+y^{2}-4 x-8 y+16=0 \text { and } \\\\ x^{2}+y^{2}-6 x-16 y+64=0 \text { can be } \end{array} $

written as

$ S_{1} \equiv(x-2)^{2}+(y-4)^{2}=2^{2} $

$ \Rightarrow \quad $ Centre $ \equiv(2,4) $ and Radius $ =2 $

$ S_{2} \equiv(x-3)^{2}+(y-8)^{2}=3^{2} $

$ \Rightarrow \quad $ Centre $ \equiv(3,8) $ and Radius $ =3 $

Equation of tangent to $ S_{1} $

$ \begin{array}{l} \Rightarrow \quad(y-4)=m(x-2)+2 \sqrt{1+m^{2}} \\\\ \Rightarrow \quad y-m x=4-2 m+2 \sqrt{1+m^{2}} . . \end{array} $

Equation of tangent to $ S_{2} $

$ \begin{array}{l} \Rightarrow \quad(y-8)=m(x-3)+3 \sqrt{1+m^{2}} \\\\ \Rightarrow \quad y-m x=8-3 m+3 \sqrt{1+m^{2}} \ldots \end{array} $

If tangent is common, then Eq. (i) and

(ii) will be same

$ 4-2 m+2 \sqrt{1+m^{2}}=8-3 m+3 \sqrt{1+m^{2}} $

$ \Rightarrow \quad m-4=\sqrt{1+m^{2}} $

On squaring both sides, we get

$ \begin{array}{l} \Rightarrow \quad m^{2}+16-8 m=1+m^{2} \\\\ \Rightarrow \quad 8 m=15 \Rightarrow m=\frac{15}{8} \end{array} $

Angle between 2 circle is $ \cos \theta=\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}} $

$ c_{1} \equiv(-1,3), r_{1}=\sqrt{1+9+6}=4 $

$ c_{2} \equiv(3,1), r_{2}=\sqrt{9+1-K}=\sqrt{10-K} $

$ c_{1} c_{2}=d=\sqrt{4^{2}+2^{2}}=\sqrt{20} $

$ \frac{-5}{24}=\frac{16+10-K-20}{2 \cdot 4 \cdot \sqrt{10-K}} $

$ \frac{-5}{3}=\frac{6-K}{\sqrt{10-K}} $

$ 5 \sqrt{10-K}=3 K-18 $

$ 250-25 K=9 K^{2}+324-108 K $

$ 9 K^{2}-83 K+74=0 $

$ 9 K^{2}-74 K-9 K+74=0 $

$ K=\frac{74}{9}, K=1 $

$ c=(p, q) $, constant $ \tan =c $

$ c_{1}=(1,2), r_{1}=\sqrt{1+4-4}=1 $

$ c_{2}=(-1,2), r_{2}=\sqrt{1+4-1}=2 $

$ c_{3}=(2,1), r_{3}=\sqrt{4+1+11}=4 $

$ 2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2} $

[ $ \because $ Condition for orthogonal]

$ \begin{array}{l} 2 p+4 q=c+4 \\\\ -2 p+4 q=c+1 \\\\ 4 p+2 q=c-11 \\\\ \text { btracting Eq: (i) fro } \\\\ 4 p=3 \Rightarrow p=\frac{3}{4} \end{array} $

On subtracting Eq. (i) from (ii), we get

On subtracting Eqs. (iii) from (ii), we get

$ -6 p+2 q=12 \Rightarrow-6 \times \frac{3}{4}+2 q=12 $

$ 2 q=12+\frac{9}{2} \Rightarrow 2 q=\frac{33}{2} \Rightarrow q=\frac{33}{4} $

$ \therefore \quad p+q=\frac{33}{4}+\frac{3}{4}=\frac{36}{4}=9 $