Solution:

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

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 $x^2+y^2+2gx+2fy+c=0$

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

Condition for orthogonality is $2{\mathrm{gg}}_1+2{\mathrm{ff}}_1=\mathrm{c}+{\mathrm{c}}_1$ $\therefore 0+2\mathrm{f}(-3)=\mathrm{C}+1$ 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 $x^2+y^2+2gx-1=0$

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

Since the line $3x+4y+5=0$ 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 $4x^2+4y^2-15x-4=0$

Solution:

Let the equation of the circles be

$x^2+y^2+2gx+2fy+d=0$

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 $y=mx+c$ 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

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

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

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

$\therefore$ product of the roots ${\mathrm{g}}_1{\mathrm{g}}_2={\mathrm{a}}^2(1+{\mathrm{m}}^2)-{\mathrm{c}}^2$

Sum of roots ${\mathrm{g}}_1+{\mathrm{g}}_2=-2\mathrm{mc}$

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

These two circles will be orthogonal if

$\begin{array}{rl}& 2{\mathrm{g}}_1{\mathrm{g}}_2=-{\mathrm{a}}^2-{\mathrm{a}}^2\\ & {\mathrm{g}}_1{\mathrm{g}}_2=-{\mathrm{a}}^2\end{array}$

But ${\mathrm{g}}_1{\mathrm{g}}_2=-{\mathrm{c}}^2+{\mathrm{a}}^2(1+{\mathrm{m}}^2)$

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

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

Which is the required condition

Solution:

Let $C_1$ and $C_2$ be the centres of given circles $C_1(\frac{-1}{2},\frac{-1}{2})$ and $C_2(\frac{-1}{2},\frac{1}{2})$

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