Circle

$ \begin{aligned} &\text { Let equation of circle be }\\ &\begin{aligned} & x^2+y^2+2 g x+2 f y=0 \\ & \Rightarrow \quad 0+\frac{100}{9}+2 f \times \frac{10}{3}=0 \\ & \Rightarrow \quad 2 f \times \frac{10}{3}=\frac{-10}{3} \times \frac{10}{3} \\ & \Rightarrow \quad f=\frac{-5}{3} \\ & \Rightarrow \quad 2^2+2^2+4 g+4 f=0 \\ & \Rightarrow \quad 8+4 g-\frac{20}{3}=0 \\ & \Rightarrow \quad 4 g=\frac{20}{3}-8=\frac{-4}{3} \Rightarrow g=-\frac{1}{3} \\ & \Rightarrow \quad r=\sqrt{g^2+f^2}=\sqrt{\frac{25}{9}+\frac{1}{9}}=\frac{\sqrt{26}}{3} \end{aligned} \end{aligned} $

$ \text { Let mid-point of chord } A B \text { be }(8, \alpha) $

Equation of chord $A B$ will be $T=S_1$

$ \begin{aligned} & 8 x+y \alpha-75=8^2+\alpha^2-75 \\ & 8 x+y \alpha=8^2+\alpha^2 \\ & \text { Slope }-\frac{-8}{\alpha} \text { and } 8^2+\alpha^2 \leq 75 \\ & \alpha^2 \leq 1 \end{aligned} $

Slope will be an integer for $\alpha= \pm 1, \pm 2$

Number of chord will be 4.

$ \text { Radius of circle } S=0 \text { is } C, P $

$ \begin{aligned} C_1 P & =\sqrt{5^2+2^2-19}=\sqrt{10} \\ C_1 & \equiv(5,2) \end{aligned} $

$ \begin{array}{ll} \Rightarrow & C_2 P=\frac{C_1 P}{2}=\frac{\sqrt{10}}{2} \\ \Rightarrow & C_1 C_2+C_2 P=C_1 P \\ \Rightarrow & C_1 C_2=\frac{C_1 P}{2} \\ \Rightarrow & C_1 C_2=C_2 P \end{array} $

Let $C_2 \equiv(\alpha, \beta)$

$C_2$ is the mid-point of $C_1 P$.

$ \alpha=\frac{5+2}{2}, \beta=\frac{2+3}{2} $

$ \begin{aligned} & \left(x-\frac{7}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=\frac{10}{4} \\ & \Rightarrow(2 x-7)^2+(2 y-5)^2=10 \\ & \Rightarrow 4 x^2+4 y^2-28 x-20 y+49+25-10=0 \\ & \Rightarrow x^2+y^2-7 x-5 y+16=0 \end{aligned} $

$ \begin{aligned} &\text { } C A \cdot C P=r^2\\ &\begin{aligned} & \Rightarrow \sqrt{(2-1)^2+2^2} \times \sqrt{\left(2-\frac{7}{5}\right)^2+\left(0-\frac{6}{5}\right)^2}=r^2 \\ & \Rightarrow \sqrt{2^2+1^2} \times \sqrt{\frac{9}{25}+\frac{36}{25}}=r^2 \\ & \Rightarrow \sqrt{5} \times \frac{\sqrt{45}}{5}=r^2 \Rightarrow r^2=3 \Rightarrow r=\sqrt{3} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let equation of circle } S=0 \text { be }\\ &\begin{aligned} & x^2+y^2+2 g x+2 f y+C=0 \\ & \Rightarrow \quad 2 g \times 2+2 f \times 0=c-7 \quad \rightarrow(\text { i }) \\ & \Rightarrow \quad 2 g \times 0+2 f \times \frac{1}{2}=c+0 \rightarrow(\text { ii }) \\ & \Rightarrow \quad 2 g \times \frac{3}{4} \times 2 f \times \frac{5}{4}=c-\frac{9}{2} \rightarrow(\text { iii }) \end{aligned} \end{aligned} $

$ \begin{array}{ll} & 4 g=c-7 \rightarrow(\mathrm{iv}) \\ & f=c \rightarrow(\mathrm{v}) \\ & 3 g+5 f=2 c-9 \rightarrow(\mathrm{vi}) \\ \Rightarrow & 3\left(\frac{c-7}{4}\right)+5 c=2 c-9 \\ \Rightarrow \quad & 3 c-21+20 c=8 c-36 \\ & 15 c=-15 \Rightarrow c=-1 \\ & g=-2, f=-1 \end{array} $

Equation of circle $s=0$

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

Equation of radical axis of $s=0$ and $s_1=0$ will be $s-s_1=0$.

$ \begin{array}{r} -8 x-2 y+6=0 \\ \Rightarrow \quad 4 x+y-3=0 \end{array} $

Equation of radical axis is

$ \begin{aligned} & s_1-s_2=0 \\ & (4 g-3) x+(4 f-8) y=0 \end{aligned} $

Centre of given circle $=(-1,-1)$

$ r=\sqrt{1^2+1^2-1}=0 $

Length of perpendicular from $(-1,-1)$ upon

$(4 g-3) x+(4 f-8) y$ equals radius of circle.

$ \begin{aligned} & \quad=\frac{|3-4 g+8-4 f|}{\sqrt{(4 g-3)^2+(8-4 f)^2}}=1 \\ & \Rightarrow 3-4 g+8-4 f \mid=\sqrt{(4 g-3)^2+(8-4)^2} \\ & \Rightarrow(3-4 g)(8-4 f)=0 \\ & \Rightarrow g=\frac{3}{4} \text { or } f=2 \end{aligned} $

Centre of circle is $O \equiv(1,-2)$

Radius of circle is $r=\sqrt{1^2+4+11}=4$

$ \begin{aligned} O D & =\sqrt{r^2-A D^2} \\ & =\sqrt{16-3}=\sqrt{13} \end{aligned} $

$O D=\sqrt{13}=$ Length of perpendicular from

$(1,-2)$ upon $2 x+3 y+k=0$

$ \begin{aligned} & \sqrt{13}=\frac{|2-6+k|}{\sqrt{13}} \\ & |k-4|=13 \\ & k-4= \pm 13 \\ & k=17 \text { and }-9 \end{aligned} $

∴ Sum of values of $k=17-9=8$

Let equation of circle be

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

and point is $(2,-1)$

$ \begin{array}{ll} \therefore & S_1=9 \\ \Rightarrow & 2^2+1^2+4 g-2 f+c=9 \\ \Rightarrow & 4 g-2 f+c=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } & -g-f=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \text { and } & g^2+f^2-c=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{array} $

$ \begin{array}{ll} \Rightarrow & g^2+f^2+4 g-2 f=20 \\ \Rightarrow & 2 g^2+4 g+2 g=20 \\ & 2 g^2+6 g=20 \\ \Rightarrow & g^2+3 g-10=0 \\ \Rightarrow & (g+5)(g-2)=9 \\ \Rightarrow & g=-5 \text { or } g=+2 \\ & -g=5 \text { or }-g=-2 \end{array} $

So, circle is $(-g,-f) \equiv(-2,2)$

So, $\beta-\alpha=4$

$ \begin{aligned} &\text { } 2 \theta=2 \tan ^{-1}\left(\frac{4}{3}\right)\\ &\tan \theta=\frac{4}{3} \end{aligned} $

$ \begin{aligned} & \sin \theta=\frac{4}{5}=\frac{r}{O P} \\ & \frac{4}{5}=\frac{\sqrt{9+9-2}}{O P} \\ \Rightarrow & O P^2=25 \\ \Rightarrow & (k+3)^2+(6 k-3)^2=25 \\ \Rightarrow & 37 k^2-30 k-7=0 \\ \Rightarrow & k=1 \end{aligned} $

$ \begin{aligned} C A & =r=\sqrt{2^2+1^2-1} \\ & =2 \end{aligned} $

$ \begin{aligned} C P & =\sqrt{4^2+2^2}=2 \sqrt{5} \\ \sin \theta & =\frac{2}{2 \sqrt{5}}=\frac{1}{\sqrt{5}} \\ \operatorname{Area}(\triangle A B C) & =\frac{1}{2} r^2 \sin (\pi-2 \theta) \\ & =\frac{1}{2} \times 4 \times 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}} \\ & =\frac{8}{5} \end{aligned} $

$ \begin{aligned} \cos \theta & =\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2} \\ d & =C_1 C_2 \\ C_1 & \equiv(2,-1), C_2=(1,-2) \\ r_1 & =\sqrt{2^2+1^2+4} \\ r_2 & =\sqrt{1^2+2^2+11} \\ r_1 & =3, r_2=4 \\ \Rightarrow \quad d^2 & =1^2+1^2=2 \\ \Rightarrow \cos \theta & =\frac{9+16-2}{2 \times 3 \times 4}=\frac{23}{24} \\ \sin \theta & =\frac{\sqrt{47}}{24} \end{aligned} $

$ \begin{aligned} &\text { Equation of circle passing through }\\ &\begin{aligned} & x+y-2=0 \text { and } \\ & x^2+y^2+2 x-4 y+4=0 \text { is } \\ & x^2+y^2+2 x-4 y+4+\lambda(x+y-2=0 \\ & S=x^2+y^2+x(2+\lambda)+y(\lambda-4)+4-2 \lambda=0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { The circle } S=0 \text { is orthogonal to }\\ &\begin{aligned} & x^2+y^2-2 x-4 y-4=0 \\ & \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\ & \Rightarrow(2+\lambda)(-1)+(\lambda-4)(-2 \lambda=4-2 \lambda-4 \\ & \Rightarrow-2-\lambda-2 \lambda+8=-2 \lambda \\ & \Rightarrow \quad \lambda=6 \\ & \text { Radius of circle } S=0=\sqrt{4^2+1^2+8} \\ & \quad=5 \end{aligned} \end{aligned} $

Given, equation of circle

$ x^2+y^2+2 g x+2 f y-23=0 $

Centre $O(3,-2) \equiv(-g,-f)$

$ \begin{aligned} & g=-3, f=2 \\ & \text { Radius }=\sqrt{g^2+f^2-c}=\sqrt{9+4+23}=6 \\ & O P=\sqrt{(3+1)^2+(-2+5)^2}=\sqrt{16+9}=5 \\ & \because r>O P \end{aligned} $

∴ Minimum distance

$ \begin{aligned} & \frac{5 \alpha+3}{6}=-1 \Rightarrow \alpha=\frac{-9}{5} \\ & \frac{5 \beta-2}{6}=-5 \Rightarrow \beta=\frac{-28}{5} \\ \therefore & A\left(\frac{-9}{5}, \frac{-28}{5}\right) \end{aligned} $

Equation of circle

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

Passing through $A(1,2)$

$ (1-a)^2+(2-a)^2=a^2 $

$ \begin{aligned} &\begin{aligned} & a^2-6 a+5=0 \\ & a=1,5 \\ & S_1 \Rightarrow(x-1)^2+(y-1)^2=1 \\ & S_2 \Rightarrow(x-5)^2+(y-5)^2=5^2 \end{aligned}\\ &\text { Equation of } A B \Rightarrow S_1-S_2=0\\ &8 x+8 y-24-24=-24 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & x+y-3=0 \\ & L:\left(\frac{5 k+1}{k+1}, \frac{5 k+1}{k+1}\right) \end{aligned}\\ &\text { Lies on } A B \text {, }\\ &\begin{gathered} 10 k+2-3 k-3=0 \\ 7 k=1 \\ k=\frac{1}{7} \Rightarrow L=\left(\frac{3}{2}, \frac{3}{2}\right) \\ \therefore \quad A B=2 A L \\ =2 \sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{2}{\sqrt{2}}=\sqrt{2} \end{gathered} \end{aligned} $

Given, equation of circle

$ S: x^2+y^2-2 x+6 y+c=0 $

Centre $O(1,-3)$

$ \begin{aligned} r & =\sqrt{1+9-c}=\sqrt{10-c} \\ O P & =\left|\frac{4+9+2}{\sqrt{4^2+3^2}}\right|=3 \end{aligned} $

$ \begin{array}{ll} \therefore & O R=\sqrt{3^2+4^2}=5 \\ \Rightarrow & \sqrt{10-c}=5 \Rightarrow c=-15 \\ & S_1(1, k)=1^2+k^2-2+6 k-15=0 \\ \Rightarrow & k^2+6 k-16=0 \\ \Rightarrow & (k+8)(k+2)=0 \Rightarrow k=-8,2 \\ \because & k>0 \\ \therefore & k=2 \end{array} $

$(\text { Normal })_1=2 x-3 y+5=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$(\text { Normal })_2=4 x-5 y+7=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ x=2, y=3 $

∵ Centre $O(2,3)$

∴ Radius $=\sqrt{(2-2)^2+(5-3)^2}=2$

Given equation of circles

$ \begin{aligned} & S_1: x^2+y^2+2 x-3 y-5=0 \\ & g_1=1, f_1=\frac{-3}{2}, c_1=-5 \\ & S_2: x^2+y^2-3 x+2 y+1=0 \\ & g_2=\frac{-3}{2}, f_2=1, c_2=1 \end{aligned} $

Let $S=x^2+y^2+2 \alpha x+2 \beta y+c=0$

$ \begin{aligned} \therefore \quad & 2 g g_1+2 f f_1=c_1+c_2 \\ & 2 \alpha-3 \beta=c-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 2 g g_2+2 f f_2=c_1+c_2 \\ & -3 \alpha+2 \beta=c+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ & (1,-1), 1+1+2 \alpha-2 \beta+c=0 \\ & 2 \alpha-2 \beta=-2-c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

By Eqs. (i) and (iii), By Eqs. (ii) and (iii),

$ \begin{array}{ll} -\beta=2 c-3 & -\alpha=-1 \\ \Rightarrow \beta=3-2 c & \Rightarrow \alpha=1 \end{array} $

By Eqs. (iii),

$ \begin{array}{ll} \Rightarrow & 2-2 \beta=-2-c \\ \Rightarrow & 2 \beta=c+4 \\ \Rightarrow & 2(3-2 c)=c+4 \\ \Rightarrow & 6-4 c=c+4 \\ \Rightarrow & 5 c=2 \Rightarrow c=\frac{2}{5} \\ \Rightarrow & \beta=3-\frac{4}{5}=\frac{11}{5} \\ & 5 \beta=11 \\ \therefore & \alpha-5 \beta=-10 \end{array} $

We have two circles given by:

$ \begin{aligned} & S_1: x^2+y^2-4 x-6 y-12=0 \\ & S_2: x^2+y^2+6 x+18 y+26=0 \end{aligned} $

We need to find a third circle that:

• Touches both $S_1$ and $S_2$ at their point of contact
• Passes through the point $(1, -1)$

Any circle that touches both $S_1$ and $S_2$ at their point of contact can be written as:

$ \left(x^2 + y^2 - 4x - 6y - 12\right) + \lambda \left(x^2 + y^2 + 6x + 18y + 26\right) = 0 $

Since the circle also passes through the point $(1, -1)$, substitute $x=1$, $y=-1$ into the equation: $ (1)^2 + (-1)^2 - 4(1) - 6(-1) - 12 + \lambda \Big[ (1)^2 + (-1)^2 + 6(1) + 18(-1) + 26 \Big] = 0 $

Calculate each term step-by-step:

• First bracket: $1 + 1 - 4 - (-6) - 12 = 2 - 4 + 6 - 12 = -8$ (Notice $-6 \times -1 = +6$)
• Second bracket: $1 + 1 + 6 - 18 + 26 = 2 + 6 - 18 + 26 = 8 - 18 + 26 = -10 + 26 = 16$

Solve for $\lambda$:

$ \lambda = \frac{8}{16} = \frac{1}{2} $

Now plug $\lambda = \frac{1}{2}$ into our general circle equation:

$ \begin{aligned} & \left(x^2 + y^2 - 4x - 6y - 12\right) + \frac{1}{2}\left(x^2 + y^2 + 6x + 18y + 26\right) = 0 \\ & x^2 + y^2 - 4x - 6y - 12 + \frac{1}{2}x^2 + \frac{1}{2}y^2 + 3x + 9y + 13 = 0 \\ & \left(1 + \frac{1}{2}\right)x^2 + \left(1 + \frac{1}{2}\right)y^2 + (-4 + 3)x + (-6 + 9)y + (-12 + 13) = 0 \\ & \frac{3}{2}x^2 + \frac{3}{2}y^2 - x + 3y + 1 = 0 \end{aligned} $

To make it simpler, multiply both sides by $2$:

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

Divide all terms by $3$ to get standard form:

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

The center of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $(-g, -f)$.

From the equation:

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

$ \begin{aligned} & \Rightarrow(x-1)^2+(y-4)^2=\frac{(2 x+3 y-1)^2}{(4+9)} \\ & \Rightarrow\left(x^2+y^2-2 x-8 y+1+16\right)(13)=4 x^2 +9 y^2+1+12 x y-6 y-4 x \\ & \Rightarrow 9 x^2+4 y^2-12 x y-22 x-98 y+220=0 \end{aligned} $

$ \begin{aligned} & L_1: x+y=0 \\ & L_2: 2 x+y-1=0 \\ & L_3: x-3 y+2=0 \end{aligned} $

Now, equation of circumcircle

$ \begin{aligned} & \lambda_1 L_1 L_2+\lambda_2 L_2 L_3+\lambda_3 L_3 L_1=0 \\ & \lambda_1(x+y)(2 x+y-1)+\lambda_2(2 x+y-1) \\ & (x-3 y+2)+\lambda_3(x-3 y+2)(x+y)=0 \end{aligned} $

Coefficient of $x^2=$ Coefficient of $y^2$

$ \begin{aligned} & 2 \lambda_1+2 \lambda_2+\lambda_3=\lambda_1-3 \lambda_2-3 \lambda_3 \\ & \lambda_1+5 \lambda_2+4 \lambda_3=0 \\ & 5 \lambda_2=-\lambda_1-4 \lambda_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Coefficient of $x y=0$

$ \begin{aligned} & 3 \lambda_1-5 \lambda_2-2 \lambda_3=0 \\ & 5 \lambda_2=3 \lambda_1-2 \lambda_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i)and (ii),

$ \begin{aligned} \therefore \quad-\lambda_1-4 \lambda_3 & =3 \lambda_1-2 \lambda_3 \\ 4 \lambda_1 & =-2 \lambda_3 \Rightarrow \lambda_3=-2 \lambda_1 \\ \frac{\lambda_1}{\lambda_3} & =\frac{-1}{2} \end{aligned} $

And $5 \lambda_2=-\lambda_1+8 \lambda_1=7 \lambda_1$

$ \begin{array}{ll} & \lambda_2=\frac{7 \lambda_1}{5} \Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{5}{7} \\ \therefore & \frac{7 \lambda_1}{\lambda_2}+\frac{\lambda_3}{\lambda_1}=7 \times \frac{5}{7}-2=5-2=3 \end{array} $

$x_{\text {intercept }}=0$

$y_{\text {intercept }}=2$

Centre lies on $y=x+1$

Let equation of circle

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

Centre $(-g,-f) \Rightarrow-f=-g+1$

$ \begin{aligned} & f=g-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & x_{\text {intercept }}=2 \sqrt{g^2-c}=0 \\ & \Rightarrow \quad g^2=c \\ & y_{\text {intercept }}=2 \sqrt{f^2-c}=2 \\ & \Rightarrow \quad f^2-c=1 \\ & \\ & \\ & \\ & \quad f^2-g^2=1 \\ & \quad(f-g)(f+g)=1 \\ & \quad \quad f+g=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because f-g=-1]\\ & \quad\therefore f=-1, \quad g=0 \end{aligned} $

∴ Centre is $(0,1)$.

Which satisfies

$ x^2+y^2-26 x-20 y+19=0 $

Let equation of tangent be

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

∴ 1 st distance from centre $=$ radius

$ \begin{aligned} & \left|\frac{3 m-4+m-2}{\sqrt{1+m^2}}\right|=2 \\ & \Rightarrow(4 m-6)^2=4\left(1+m^2\right) \\ & \Rightarrow 3 m^2-12 m+8=0 \\ & \therefore \sqrt{3}\left|m_1-m_2\right|=\sqrt{3} \frac{\sqrt{D}}{a} \\ & =\sqrt{3} \frac{\sqrt{144-4 \times 8 \times 3}}{3} \\ & =\frac{4}{\sqrt{3}} \times \sqrt{3}=4 \end{aligned} $

$ \begin{array}{ll} \because & P A=P B \\ \therefore & \triangle O A P \cong \triangle O P B . \end{array} $

( $O$ is centre of circle

$ \begin{aligned} & x^2+y^2-4 x-4 y-8=0 \\ & r=\sqrt{4+4+8}=4) \end{aligned} $

∵ Equation of circle at $P$ and radius 4 is

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

$\therefore A B$ is common chord

∴ Equation is $S_1-S_2=0$

$ \begin{aligned} &\left(x^2+y^2-4 x-4 y-8\right) \\ & -\left(x^2+y^2-4 x+4 y+4+4-4\right)=0 \\ & \Rightarrow \quad-8 y-8-4=0 \\ & \Rightarrow \quad 8 y+12=0 \\ & 2 y+3=0 \end{aligned} $

∵ If two circles cut orthogonally.

$ \begin{aligned} & \Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1 c_2 \\ & S_1: x^2+y^2-2 y-3=0 \\ & S_2: x^2+y^2+4 x+3=0 \\ & S_3: x^2+y^2-2 \alpha x-2 \beta x+c=0 \rightarrow \text { centre } \\ & (\alpha, \beta) \\ & \therefore 2(-\alpha)(0)+2(-\beta)(-1)=c-3 \\ & \quad 2 \beta=c-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

And $2(-\alpha)(2)+2 \beta(0)=c+3$

$ -4 \alpha=c+3 \Rightarrow 4 \alpha=-c-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On adding Eqs. (i) and (ii), we get

$ \begin{aligned} & 4 \alpha+2 \beta=-6 \\ \Rightarrow \quad & 2 \alpha+\beta=-3 \end{aligned} $

Circle $C_1 \rightarrow$ radius $r_1, O_1(1,2), r_1=3 r^{\prime}$

Circle $\mathrm{C}_2 \rightarrow$ radius $r_2, \mathrm{O}_2(3,-2), r_2=r^{\prime}$

$\because C_1$ and $C_2$ cuts orthogonally

$ \begin{aligned} & r_1^2+r_2^2=0_1 0_2^2 \\ \Rightarrow \quad & \left(3 r^{\prime}\right)^2+r^{\prime 2}=2^2+4^2 \\ \Rightarrow \quad & 10 r^{\prime 2}=20 \Rightarrow r^{\prime}=2 \end{aligned} $

∴ Equation of required circle

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

Given, equation of circles

$ x^2+y^2=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) $

And $(x-9)^2+y^2=16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$ \begin{aligned} C_1 & =(0,0), r_1=4 \\ C_2 & =(9,0), r_2=4 \\ \therefore \quad C_1 C_2 & =\sqrt{(9-0)^2+(0-0)^2} \\ & =\sqrt{(9)^2}=9 \end{aligned} $

And $r_1+r_2=4+4=8$

Since $C_1 C_2>r_1+r_2$

Circles are separate to each other, any tangent to circle (i) with slope ( $m$ ) is

$ \begin{aligned} & y=m x \pm r \sqrt{1+m^2} \\ \Rightarrow \quad & y=m x+4 \sqrt{1+m^2} \\ \Rightarrow \quad & m x-y \pm 4 \sqrt{1+m^2}=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Equation (iii) to be tangent to circle (ii)

$ \begin{array}{ll} \therefore & \left|\frac{m(9)-0 \pm 4 \sqrt{1+m^2}}{\sqrt{m^2+1}}\right|=4 \\ \Rightarrow & 16\left(m^2+1\right)=\left(9 m \pm 4 \sqrt{1+m^2}\right)^2 \\ \Rightarrow & 16 m^2+16=81 m^2+16\left(1+m^2\right) \\ & \pm 72 m \sqrt{1+m^2} \end{array} $

$ \begin{array}{ll} \Rightarrow & \quad 81 m^2 \pm 72 m \sqrt{1+m^2}=0 \\ \Rightarrow & 9 m\left(9 m \pm 8 \sqrt{1+m^2}\right)=0 \end{array} $

When $9 m=0 \Rightarrow m=0$

Then Eq. (iii) $y= \pm 4$ (horizontal lines)

If, $9 m \pm 8 \sqrt{1+m^2}=0$

$ \begin{aligned} & \Rightarrow \quad(9 m)^2=\left( \pm 8 \sqrt{1+m^2}\right)^2 \\ & \Rightarrow \quad 81 m^2=64\left(1+m^2\right) \\ & \Rightarrow \quad 81 m^2-64 m^2=64 \\ & \Rightarrow \quad 17 m^2=64 \\ & \Rightarrow \quad m= \pm \sqrt{\frac{64}{17}} \\ & \Rightarrow \quad m= \pm \frac{8}{\sqrt{17}} \end{aligned} $

$ \begin{aligned} &\text { Given, equation of circle }\\ &\begin{aligned} & x^2+y^2-4 x-6 y-12=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { centre }(-g,-f)=(2,3) \\ & \text { And radius }=\sqrt{g^2+f^2-c} \\ & \qquad \begin{aligned} & =\sqrt{4+9+12}=\sqrt{25}=5 \end{aligned} \end{aligned} \end{aligned} $

Equation of new circle whose radius is 3

$ \begin{aligned} & (x-h)^2+(y-k)^2=(3)^2 \\ & \Rightarrow \quad(x-h)^2+(y-k)^2=9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Since circle (i) and (ii) touches each other internally at $(-1,-1)$

$ \therefore \quad C_1 C_2=\left|r_1-r_2\right| $

where $c_2=(h, k)$

$ \therefore \quad C_1 C_2=|5-3|=2 $

$P$ divides the line segment $C_1 C_2$ in the ratio $r_1: r_2$

$ \quad \begin{aligned}\therefore C_1 P & =r_1 \text { and } C_2 P=r_2 \\ C_1 P & =\sqrt{(-1-2)^2+(-1-3)^2} \\ & =\sqrt{(-3)^2+(-4)^2} \\ & =\sqrt{9+16}=\sqrt{25}=5 \end{aligned} $

and $C_2 P=r_2=3$

$\therefore$ The point of tangency $P$ divides the line segment $C_1 C_2$ externally in the ratio $r_1:-r$

$ \begin{array}{rlrl} \therefore & P =\frac{r_1 C_2-r C_1}{r_1-r} \\ & =\frac{5 C_2-3 C_1}{5-3}=\frac{5 C_2-3 C_1}{2} \\ \Rightarrow & 2 P =5 C_2-3 C_1 \\ \Rightarrow & 5 C_2 =2 P+3 C_1 \\ \Rightarrow & 5(h, k) =2(-1,-1)+3(2,3) \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 5(h, k)=(-2,-2)+(6,9) \\ & \Rightarrow \quad 5(h, k)=(4,7) \Rightarrow(h, k)=\left(\frac{4}{5}, \frac{7}{5}\right) \end{aligned}\\ &\therefore \text { Equation of required circle }\\ &\begin{aligned} & \left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=(3)^2 \\ & \Rightarrow \quad\left(x-\frac{4}{5}\right)^2+\left(y-\frac{7}{5}\right)^2=9 \\ & \Rightarrow \quad x^2+\frac{16}{25}-\frac{8}{5} x+y^2+\frac{49}{25}-\frac{14}{5} y=9 \\ & \quad x^2+y^2-\frac{8}{5} x-\frac{14}{5} y+\frac{13}{5}=9 \\ & \Rightarrow \quad 5 x^2+5 y^2-8 x-14 y-32=0 \end{aligned} \end{aligned} $

If $C_1$ and $C_2$ are two circles having no common points, it means they do not intersect and do not touch each other then there will be no common tangents or there will be four common tangents $C_1$ and $C_2$

Equation of the circle touching the $X$-axis and passing through the point $(-1,1)$

$ \begin{array}{ll} & (-1-h)^2+(1-k)^2=k^2 \\ \Rightarrow & 1+h^2+2 h+1+k^2-2 k=k^2 \\ \Rightarrow & 1+2 h+h^2+1-2 k+k^2=k^2 \\ \Rightarrow & 1+2 h+h^2+1-2 k=0 \\ \Rightarrow & h^2+2 h-2 k+2=0 \\ \Rightarrow & x^2+2 x-2 y+2=0 \\ \Rightarrow & 2 y=x^2+2 x+2 \\ \Rightarrow & y=\frac{1}{2}\left(x^2+2 x\right)+1 \\ \Rightarrow & y=\frac{1}{2}(x+1)^2-\frac{1}{2}+1 \\ \Rightarrow & y=\frac{1}{2}(x+1)^2+\frac{1}{2} \\ \Rightarrow & (x+1)^2=2\left(y-\frac{1}{2}\right) \end{array} $

Which represents parabola with focus $(-1,1)$

Given, equation of circles

$ \begin{aligned} & x^2+y^2+2 x-2 y+1=0 \text { and } \\ & x^2+y^2-2 x+2 y-2=0 \end{aligned} $

$\therefore$ Equation of circle passing through the point of intersection of two given circle is

$ \begin{array}{r} \left(x^2+y^2+2 x-2 y+1\right)+\lambda \left(x^2+y^2-2 x+2 y-2\right)=0 \\ \Rightarrow \quad(1+\lambda) x^2+(1+\lambda) y^2+(2-2 \lambda) x +(-2+2 \lambda) y+(1-2 \lambda)=0 \\ \Rightarrow \quad x^2+y^2+\frac{(2-2 \lambda)}{1+\lambda} x+\frac{(-2+2 \lambda)}{1+\lambda} y +\frac{1-2 \lambda}{1+\lambda}=0 \end{array} $

Coordinate of centre of circle

$ =(-g,-f)=\left(-\frac{1-\lambda}{1+\lambda},-\frac{-1+\lambda}{1+\lambda}\right) $

$ \begin{aligned} & \therefore \text { Radius of } \\ & \begin{aligned} &\left(-\frac{1-\lambda}{1+\lambda}\right)^2+\left(-\frac{-1+\lambda}{1+\lambda}\right)^2-\left(\frac{1-2 \lambda}{1+\lambda}\right)=14 \\ & \Rightarrow \quad \frac{(1-\lambda)^2}{(1+\lambda)^2}+\frac{(\lambda-1)^2}{(1+\lambda)^2}-\frac{1-2 \lambda}{1+\lambda}=14 \\ & \Rightarrow \quad(1-\lambda)^2+(\lambda-1)^2-(1+\lambda)(1-2 \lambda) \\ &=14(1+\lambda)^2 \\ & \Rightarrow \quad 2(\lambda-1)^2-\left(1-2 \lambda+\lambda-2 \lambda^2\right) \\ &=14\left(1+\lambda^2+2 \lambda\right) \\ & \Rightarrow \quad 2\left(\lambda^2+1-2 \lambda-\left(1-\lambda-2 \lambda^2\right)\right. \\ &=14\left(1+\lambda^2+2 \lambda\right) \\ & \Rightarrow \quad 4 \lambda^2+1-3 \lambda=14+14 \lambda^2+28 \lambda \\ & \Rightarrow \quad 10 \lambda^2+31 \lambda+13=0 \end{aligned} \end{aligned} $

$\Rightarrow$ This is a quadrate equation is $\lambda$ the existence of real values for $\lambda$ conforms that such circle exists.

Coordinates of centre

$ \begin{array}{ll} & h=\frac{\lambda-1}{1+\lambda}, k=-\frac{\lambda-1}{1+\lambda} \\ \Rightarrow & h=-k \Rightarrow h+k=0 \\ \Rightarrow & x+y=0 \end{array} $

$ \begin{aligned} &\text { Given equation of circle }\\ &x^2+y^2-14 x+6 y+33=0 \ldots \text { (i) } \end{aligned} $

$ \begin{aligned} & \text { Centre of circle }=(-g,-f) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =(7,-3) \\ & \text { and radius of circle }=\sqrt{g^2+f^2-c} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{49+9-33} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sqrt{25}=5 \end{aligned} $

Since, circle cuts the $X$-axis, $y=0$

$ \begin{array}{ll} & x^2-14 x+33=0 \\ \Rightarrow & x^2-11 x-3 x+33=0 \\ \Rightarrow & x(x-11)-3(x-11)=0 \\ \Rightarrow & (x-3)(x-11)=0 \\ \Rightarrow & x=3,11 \\ \therefore & A=(3,0) \text { and } B=(11,0) \end{array} $

Since, $C$ is the mid-point of $A B$

$ \begin{aligned} \therefore \quad C & =\left(\frac{3+11}{2}, \frac{0+0}{2}\right) \\ & =(7,0) \end{aligned} $

$\therefore$ Equation of line $L$, which passes through $C$ and having slope $(-1)$ is

$ \begin{aligned} & y-0=(-1)(x-7) \\ & \Rightarrow \quad y=-x+7 \\ & \Rightarrow \quad x+y-7=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Since, $L$ is the radical axis of the circles $s$ and $S^{\prime}\left(S-S^{\prime}=0\right)$

$ \begin{aligned} & \therefore \quad\left(x^2+y^2-14 x+6 y+33\right) -\left(x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}\right)=0 \\ & \Rightarrow\left(-14-2 g^{\prime}\right) x+\left(6-2 f^{\prime}\right) y+\left(33-c^{\prime}\right)=0 \end{aligned} $

Which is identical with line (ii)

$ \begin{array}{ll} \therefore & \frac{-14-2 g^{\prime}}{1}=\frac{6-2 f^{\prime}}{1}=\frac{33-c^{\prime}}{-7}=k \\ \Rightarrow & -14-2 g^{\prime}=k \text { and } 6-2 f^{\prime}=k \\ \therefore & -14-2 g^{\prime}=6-2 f^{\prime} \\ \Rightarrow & g^{\prime}-f^{\prime}=-10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Since, centre of $S^{\prime}$ lies on (ii)

$ \begin{aligned} \therefore & & -g^{\prime}-f^{\prime} & =7 \\ \Rightarrow & & g^{\prime}+f^{\prime} & =-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Solving Eqs. (iii) and (iv), we get

$ \begin{array}{ll} & g^{\prime}=\frac{-17}{2}, f^{\prime}=\frac{3}{2} \\ \therefore & -14-2\left(-\frac{17}{2}\right)=k \\ \Rightarrow & k=3 \end{array} $

Also, $\frac{33-c^{\prime}}{-7}=k=3$

$ \begin{array}{ll} \Rightarrow & 33-c^{\prime}=-21 \\ \Rightarrow & c^{\prime}=54 \end{array} $

$\therefore$ Equation of required circle is

$ \begin{aligned} & x^2+y^2+2\left(\frac{-17}{2}\right) x+2\left(\frac{3}{2}\right) y+54=0 \\ & \Rightarrow x^2+y^2-17 x+3 y+54=0 \end{aligned} $

Given equation is

$ a x^2+a y^2+2 g x+2 f y-2=0 $

Put $(-1,0)$ into this equation, we get

$ \begin{aligned} & a(-1)^2+a(0)^2+2 g(-1)+2 f(0)-2=0 \\ & \Rightarrow \quad a-2 g-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Put $(-1,1)$, we get

$ \begin{aligned} & a(-1)^2+a(1)^2+2 g(-1)+2 f(1)-2=0 \\ & \Rightarrow \quad a+a-2 g+2 f-2=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 2 a-2 g+2 f-2=0 \\ & \Rightarrow \quad a-g+f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Put (1, 1), we get

$ \begin{aligned} & a(1)^2+a(1)^2+2 g(1)+2 f(1)-2=0 \\ & \Rightarrow \quad a+g+f-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Subtract Eq. (ii) from Eq. (iii),

$ \begin{aligned} & (a+g+f-1)-(a-g+f-1)=0 \\ & \Rightarrow \quad 2 g=0 \\ & \Rightarrow \quad g=0 \end{aligned} $

Put $g=0$ into Eq. (i), we get

$ \begin{aligned} & & a-2(0)-2 & =0 \\ \Rightarrow & & a & =2 \end{aligned} $

Given circle, $x-2=5 \cos \theta$, $y+1=5 \sin \theta$, where $\theta$ is the perimeter.

$ \begin{aligned} & \Rightarrow(x-2)^2+(y+1)^2=(5 \cos \theta)^2+(5 \sin \theta)^2 \\ & =25\left(\cos ^2 \theta+\sin ^2 \theta\right)=25 \end{aligned} $

$\therefore$ The circle has a centre at $(h, k)=(2,-1)$ and a radius, $r_c=5$

Now, the line equation is $x=1+\frac{r}{2}$ and $y=-2+\frac{\sqrt{3}}{2} r$. This is parametric form.

Let $x_1=1, y_1=-2$

The direction vector of the line is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

So, the slope of the line, $m=\frac{\Delta y}{\Delta x}$

$ \Rightarrow \quad \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} $

Now, the equation of line is

$ \begin{array}{cc} & y-y_1=m\left(x-x_1\right) \\ \Rightarrow & y-(-2)=\sqrt{3}(x-1) \\ \Rightarrow & y+2=\sqrt{3} x-\sqrt{3} \\ \Rightarrow & \sqrt{3} x-y-(2+\sqrt{3})=0 \end{array} $

Now, calculate the distance from the center of the circle to line,

$ d=\frac{\left|A x_0+B y_0+C\right|}{\sqrt{A^2+B^2}} $

Here, $\left(x_0, y_0\right)=(2,-1)$ and the line is $\sqrt{3} x-y-(2+\sqrt{3})=0$

So, $A=\sqrt{3}, B=-1, C=-(2+\sqrt{3})$

$ \begin{aligned} d & =\frac{|\sqrt{3}(2)+(-1)(-1)-(2+\sqrt{3})|}{\sqrt{(\sqrt{3})^2+(-1)^2}} \\ & =\frac{|2 \sqrt{3}+1-2-\sqrt{3}|}{\sqrt{3+1}} \\ & =\frac{\sqrt{3}-1}{2} \approx 0.366<5 \end{aligned} $

So, the line intersects the circle at two distinct points, meaning it is a secant to the circle.

So, the line is a chord of circle other than diameter.

The equation of the circle is

$ \begin{aligned} & x^2+y^2-6 x+2 y+c=0 \\ \Rightarrow \quad & (x-3)^2+(y+1)^2=10-c \end{aligned} $

So, the center of the circle is $C(3,-1)$ and the radius is $r^2=10-C$

Now, since the equation of tangent is

$ x-2 y=0 $

Slop, $m_t=1 / 2$

Slope of the radius, connecting the centre $(3,-1)$ to the point of tangency

$ P\left(x_p, y_p\right) \text { is } m_r=\frac{y_p-(-1)}{x_p-3}=\frac{y_p+1}{x_p-3} $

Now, since radius is perpendicular to the tangent, so

$ \begin{aligned} & m_t \cdot m_r=-1 \\ \Rightarrow \quad & \frac{1}{2} \cdot \frac{y_p+1}{x_p-3}=-1 \Rightarrow y_p+1=-2 x_p+6 \\ \Rightarrow \quad & y_p=-2 x_p+5 \end{aligned} $

But, $P$ lies on the tangent line, its coordinates satisfy $x_p-2 y_p=0$

$ \begin{array}{ll} \Rightarrow & x_p=2 y_p \\ \text { So, } & y_p=-2 x_p+5 \\ \Rightarrow & -2\left(2 y_p\right)+5=-4 y_p+5 \\ \Rightarrow & 5 y_p=5 \Rightarrow y_p=1 \end{array} $

So, $x_p=2 y_p=2 \times 1=2$

So, the point $P$ is $(2,1)$.

Now, the distance between $(6,3)$ and $(2,1)$ is

$ \begin{aligned} & d=\sqrt{(2-6)^2+(1-3)^2} \\ \Rightarrow & \sqrt{(-4)^2+(-2)^2} \Rightarrow \sqrt{16+4}=\sqrt{20}=2 \sqrt{5} \end{aligned} $

$\therefore$ The distance of the point $(6,3)$ from $P$ is $2 \sqrt{5}$.

The equation of the circle is

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

Comparing this with the general form

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

We have

$ \begin{aligned} & & 2 g & =-4 \\ \Rightarrow & & g & =-2 \text { and } 2 f=2 \\ \Rightarrow & & f & =1 \end{aligned} $

So, the center of the circle $C$ is

$ (-g,-f)=(2,-1) $

And, the radius is $r=\sqrt{g^2+f^2-c}$

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

The point from which tangents are drawn is $P=(-3,1)$

Now, the circumcircle of $\triangle P A B$ is the circle with $P C$ as its diameter. The center of this circumcircle is the mid-point of $P C$.

Mid-point $M=\left(\frac{-3+2}{2}, \frac{1+(-1)}{2}\right)$

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

Distance $P C=\sqrt{(2-(-3))^2+(-1-1)^2}$

$ \begin{aligned} & \Rightarrow \quad \sqrt{5^2+(-2)^2} \\ & \Rightarrow \quad \sqrt{25+4}=\sqrt{29} \end{aligned} $

$\therefore$ The radius of the circumcircle

$=\frac{1}{2} \cdot$ distance $P C=\frac{\sqrt{29}}{2}$

Now, the equation of a circle with center ( $h, k$ ) and radius $R$ is

$ (x-h)^2+(y-k)^2=R^2 $

Here, $(h, k)=\left(\frac{-1}{2}, 0\right)$ and $R=\frac{\sqrt{29}}{2}$

So, $\left(x-\left(-\frac{1}{2}\right)\right)^2+(y-0)^2=\left(\frac{\sqrt{29}}{2}\right)^2$

$ \begin{aligned} & \Rightarrow \quad\left(x+\frac{1}{2}\right)^2+y^2=\frac{29}{4} \\ & \Rightarrow \quad x^2+x+\frac{1}{4}+y^2=\frac{29}{4} \\ & \Rightarrow \quad x^2+y^2+x-\frac{28}{4}=0 \\ & \Rightarrow \quad x^2+y^2+x-7=0 \end{aligned} $

Let the equation of circle $C$ be

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

Since, it passes through $(1,1)$ then

$ \begin{array}{r} 1^2+1^2+2 g+2 f+c=0 \\ \Rightarrow \quad 2+2 g+2 f+c=0         \,\,\,...(i)\end{array} $

Now, $S_1: x^2+y^2+2 g x+2 f y+c=0$

and $S_2=x^2+y^2-2 x=0$

So, $S_1-S_2=0$ is the common chord.

$ \Rightarrow(2 g+2 x+2 f y+c=0 $

Now, center of $s_2=x^2+y^2-2 x=0$ is $(1,0)$

Since, the common chord is a diameter of $S_2$, then centre of $S_2$ must lies on the common chord.

$ \begin{array}{ll} \therefore & (2 g+2(l)+2 f(0)+c=0 \\ \Rightarrow & 2 g+2+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Also, circle $C$ is orthogonal to

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

Here, $g_1=g, f_1=f, c_1=c$ and $g_2=0$,

$ f_2=1, c_2=-3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Using the orthogonality condition

$ \begin{aligned} & 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\ \Rightarrow \quad & 2 g(0)+2 f(1)=c+(-3) \\ \Rightarrow \quad & 2 f=c-3 \end{aligned} $

From Eq. (ii), $c=-2 g-2$

Put this in Eq. (iii), we get

$ \begin{aligned} & 2 f=c-3=-2 g-2-3 \\ \Rightarrow \quad & 2 f=-2 g-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Substitute $c=-2 g-2$ into Eq. (i), we get

$ \begin{array}{ll} \Rightarrow & 2+2 g+2 f-2 g-2=0 \\ \Rightarrow & 2 f=0 \\ \Rightarrow & f=0 \end{array} $

Put $f=0$ in Eq. (iv), we get

$ \begin{aligned} & & 2(0) & =-2 g-5 \\ \Rightarrow & & 2 g & =-5 \\ \Rightarrow & & g & =\frac{-5}{2} \end{aligned} $

$ \begin{aligned} &\text { The center of circle } C \text { is }\\ &\begin{aligned} & (-g,-f)=\left(-\left(-\frac{5}{2}\right), 0\right) \\ \Rightarrow & \left(\frac{5}{2}, 0\right) \end{aligned} \end{aligned} $

For $S_1: x^2+y^2-2 x+k y+1=0$,

Center $C_1=\left(1,-\frac{k}{2}\right)$

Radius, $r_1=\sqrt{1^2+\left(-\frac{k}{2}\right)^2-1}=\sqrt{\frac{k^2}{4}}=\frac{k}{2}$

For $S_2: x^2+y^2-k x-2 y+1=0$

Center $C_2=\left(\frac{k}{2}, 1\right)$

Radius, $r_2=\sqrt{\left(\frac{k}{2}\right)^2+1^2-1}$

$ \sqrt{\frac{k^2}{4}}=\frac{k}{2} $

Since, angle $\theta$ between two circles is given by $\cos \theta=\frac{d^2-r_1^2-r_2^2}{2 r_1 r_2}$, where $d$ is the distance between the centers.

Given, $\theta=\cos ^{-1}\left(\frac{1}{4}\right)$

$ \begin{aligned} \Rightarrow \quad \cos \theta & =\frac{1}{4} \\ d^2 & =\left(1-\frac{k}{2}\right)^2+\left(\frac{-k}{2}-1\right)^2 \\ & =1-k+\frac{k^2}{4}+1+k+\frac{k^2}{4} \\ & =2+\frac{k^2}{2} \end{aligned} $

Since $k<0, r_1=\frac{-k}{2}$ and $r_2=\frac{-k}{2}$

Substituting into the formula

$ \frac{1}{4}=\frac{\left(2+\frac{k^2}{2}\right)-\left(\frac{-k}{2}\right)^2-\left(\frac{-k}{2}\right)^2}{2\left(\frac{-k}{2}\right)\left(\frac{-k}{2}\right)} $

$ \begin{array}{ll} \Rightarrow & \frac{1}{4}=\frac{2+\frac{k^2}{2}-\frac{k^2}{4}-\frac{k^2}{4}}{2\left(\frac{k^2}{4}\right)} \\ \Rightarrow & \frac{1}{4}=\frac{4}{k^2} \\ \Rightarrow & k^2=16 \\ \Rightarrow & k= \pm 4 \end{array} $

But since $k<0, k=-4$

Now, the equation of the radical axis is

$ \begin{aligned} & S_1-S_2=0 \\ & \left(x^2+y^2-2 x+k y+1\right)-\left(x^2+y^2-k x\right. \\ & -2 y+1)=0 \\ & \Rightarrow-2 x+k y+k x+2 y=0 \\ & \Rightarrow-6 x-2 y=0 \,\,\,\,(\because k=-4)\\ & \Rightarrow 3 x+y=0 \end{aligned} $

The points satisfying this equation are $(-1,3)$ and $(1,-3)$ lies on the radical axis.

$ 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 $