Solution:

Here $S \equiv x^{2}+y^{2}-4 x+6 y+11=0$ and $S^{\prime}=x^{2}+y^{2}-2 x+8 y+13=0$

The centre of these circles are $\mathrm{C} _{1}(2,-3)$ and $\mathrm{C} _{2}(1,-4)$ respectively.

The radius of these circles are $r _{1}=\sqrt{4+9-11}=\sqrt{2}$

and $\mathrm{r} _{2}=\sqrt{1+16-13}=2$

Distance between the centres $\mathrm{C} _{1} \mathrm{C} _{2}=\mathrm{d}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$

Angle between two circle is $\cos \theta=\left|\frac{\mathrm{r} _{1}{ }^{2}+\mathrm{r} _{2}{ }^{2}-\left(\mathrm{c} _{1} \mathrm{c} _{2}\right)^{2}}{2 \mathrm{r} _{1} \mathrm{r} _{2}}\right|=\left|\frac{2+4-2}{2 \times 2 \times \sqrt{2}}\right|=\frac{1}{\sqrt{2}}=\cos \frac{\pi}{4}$

$\therefore \theta=\frac{\pi}{4}$

Solution:

Let the required equation of circle be $x^{2}+y^{2}+2 g x+2 f y+c=0$

This circle intersect orthogonally with circles $x^{2}+y^{2}-6 y+1=0$ and $x^{2}+y^{2}-4 y+1=0$

Condition for orthogonality is $2 \mathrm{gg} _{1}+2 \mathrm{ff} _{1}=\mathrm{c}+\mathrm{c} _{1}$ $\therefore 0+2 \mathrm{f}(-3)=\mathrm{C}+1 \quad$ and $0+2 \mathrm{f}(-2)=\mathrm{c}+1$ $-6 \mathrm{f}=\mathrm{c}+1$ $-4 \mathrm{f}=\mathrm{c}+1$

$\therefore-6 \mathrm{f}=-4 \mathrm{f} \Rightarrow \mathrm{f}=0$ and $\mathrm{c}=-1$

Equation of circle is $x^{2}+y^{2}+2 g x-1=0$

Centre is $(-\mathrm{g}, 0)$ and radius $\sqrt{\mathrm{g}^{2}+1}$

Since the line $3 x+4 y+5=0$ touches the circle

$\therefore$ distance of this line from the centre must be equal to radius $\sqrt{\mathrm{g}^{2}+1}$

$\therefore\left|\frac{-3 \mathrm{~g}+5}{\sqrt{9+16}}\right|=\sqrt{\mathrm{g}^{2}+1}$

$5-3 \mathrm{~g}=5 \sqrt{\mathrm{g}^{2}+1}$

squaring

$25+9 \mathrm{~g}^{2}-30 \mathrm{~g}=25 \mathrm{~g}^{2}+25$

$16 \mathrm{~g}^{2}+30 \mathrm{~g}=0$

$2 \mathrm{~g}(8 \mathrm{~g}+15)=0$

$g=0$ or $g=\frac{-15}{8}$

Hence equations of circles are

$\mathrm{x}^{2}+\mathrm{y}^{2}-1=0$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-\frac{15}{4} \mathrm{x}-1=0$

$x^{2}+y^{2}-1=0$ and $4 x^{2}+4 y^{2}-15 x-4=0$

Solution:

Let the equation of the circles be

$x^{2}+y^{2}+2 g x+2 f y+d=0$

This circle passes through the points $(0, \mathrm{a})$ and $(0,-\mathrm{a})$

$\therefore \mathrm{a}^{2}+2 \mathrm{fa}+\mathrm{d}=0$

(1) and $\mathrm{a}^{2}-2 \mathrm{fa}+\mathrm{d}=0$

(1) (2)

$4 \mathrm{fa}=0$

$\therefore \mathrm{f}=0$ and $\mathrm{d}=-\mathrm{a}^{2}$

$\therefore$ The equation of circle is $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}-\mathrm{a}^{2}=0$

Centre of this circle is $(-\mathrm{g}, 0)$ and radies $\sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}$

Since line $y=m x+c$ touches the circle

$\therefore\left|\frac{-\mathrm{mg}+\mathrm{c}}{\sqrt{\mathrm{m}^{2}+1}}\right|=\sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}}$

$\mathrm{c}-\mathrm{mg}=\sqrt{\mathrm{g}^{2}+\mathrm{a}^{2}} \sqrt{\mathrm{m}^{2}+1}$

Squaring

$\mathrm{c}^{2}+\mathrm{m}^{2} \mathrm{~g}^{2}-2 \mathrm{mcg}=\mathrm{g}^{2} \mathrm{~m}^{2}+\mathrm{g}^{2}+\mathrm{a}^{2} \mathrm{~m}^{2}+\mathrm{a}^{2}$

$\mathrm{g}^{2}+2 \mathrm{mcg}+\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)-\mathrm{c}^{2}=0$

It is a quadratic in $\mathrm{g}$

$\therefore$ product of the roots $\mathrm{g} _{1} \mathrm{~g} _{2}=\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)-\mathrm{c}^{2}$

Sum of roots $\mathrm{g} _{1}+\mathrm{g} _{2}=-2 \mathrm{mc}$

Now the equations of the two circles represented are $x^{2}+y^{2}+2 g _{1} x-a^{2}=0$ and $x^{2}+y^{2}+2 g _{2} x-a^{2}=$ 0

These two circles will be orthogonal if

$ \begin{aligned} & 2 \mathrm{~g} _{1} \mathrm{~g} _{2}=-\mathrm{a}^{2}-\mathrm{a}^{2} \\ & \mathrm{~g} _{1} \mathrm{~g} _{2}=-\mathrm{a}^{2} \end{aligned} $

But $\mathrm{g} _{1} \mathrm{~g} _{2}=-\mathrm{c}^{2}+\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)$

$\therefore-\mathrm{c}^{2}+\mathrm{a}^{2}\left(1+\mathrm{m}^{2}\right)=-\mathrm{a}^{2}$

or $\mathrm{c}^{2}=\mathrm{a}^{2}\left(2+\mathrm{m}^{2}\right)$

Which is the required condition

Solution:

Let $C _{1}$ and $C _{2}$ be the centres of given circles $C _{1}\left(\frac{-1}{2}, \frac{-1}{2}\right)$ and $C _{2}\left(\frac{-1}{2}, \frac{1}{2}\right)$

Also radius these two circles are $r _{1}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}$

and $\mathrm{r} _{2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}$

$\cos \theta=\frac{\mathrm{r} _{1}{ }^{2}+\mathrm{r} _{2}{ }^{2}-\mathrm{d}^{2}}{2 \mathrm{r} _{1} \mathrm{r} _{2}}$

$=\frac{\frac{1}{2}+\frac{1}{2}-1}{2 \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}}$

$=0$

$\therefore \theta=\frac{\pi}{2}$

$\therefore$ Required line is parallel to $\mathrm{x}$-axis and it passes through $(1,2)$

$\therefore$ Equation of line is $\mathrm{y}=2$.

Solution:

Here $\mathrm{OA}=2$ radius of circle $\mathrm{x}^{2}+\mathrm{y}^{2}=4$ with centre $(0,0)$

The distance of $(0,0)$ from $x+y=5 \sqrt{2}$ is

$\left|\frac{-5 \sqrt{2}}{\sqrt{2}}\right|=5$

$\therefore$ The radius of the smallest circle $=\frac{5-2}{2}=\frac{3}{2}$

and $\mathrm{OC}=2+\frac{3}{2}=\frac{7}{2}$

The slope of $\mathrm{OA}=1=\tan \theta$ $\therefore \cos \theta=\frac{1}{\sqrt{2}}$ and $\sin \theta=\frac{1}{\sqrt{2}}$

$\therefore$ Centre $(0+\mathrm{OC} \cos \theta, O+O C \sin \theta)=\left(\frac{7}{2 \sqrt{2}}, \frac{7}{2 \sqrt{2}}\right)$