Solution :

Equation of tangent of $x^{2}+y^{2}-2 a x=0 a t(a(1+\cos \alpha)$,asin $\alpha)$ is

$\mathrm{ax}(1+\cos \alpha)+\mathrm{aysin} \alpha-\mathrm{a}(\mathrm{x}+\mathrm{a}(1+\cos \alpha))=0$

$\mathrm{ax}+\mathrm{axcos} \alpha+\mathrm{aysin} \alpha-\mathrm{ax}-\mathrm{a}^{2}(1+\cos \alpha)=0$

$\operatorname{axcos} \alpha+\mathrm{aysin} \alpha=\mathrm{a}^{2}(1+\cos \alpha)$

$\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{a}(1+\cos \alpha)$

Solution :

Since tangents make an angle of $60^{\circ}$ with the $\mathrm{x}$-axis so slope of tangent

$\mathrm{m}=\tan 60^{\circ}=\mathrm{m}=\sqrt{3}$

radius of circle $\mathrm{x}^{2}+\mathrm{y}^{2}=9$ is 3

we know equation of tangent to a circle $x^{2}+y^{2}=a^{2}$ is

$y=m x \pm a \sqrt{1+m^{2}}$

$\therefore \mathrm{y}=\sqrt{3} \mathrm{x} \pm 3 \sqrt{1+3}$

$=\sqrt{3} \mathrm{x} \pm 6$

or $\sqrt{3} x-y \pm 6=0$ ie., $\sqrt{3} x-y+6=0$ and $\sqrt{3} x-y-6=0$ are equations of tangents.

Solution :

Since the line $(\mathrm{x}-2) \cos \theta+(\mathrm{y}-2) \sin \theta=1$ (1) touches a circle so it is a tangent equation to a circle.

Equation of tangent to a circle at $\left(x _{1}, y _{1}\right)$ is $(x-h) x _{1}+(y-k) y _{1}=a^{2}$ to a circle $(x-h)^{2}+(y-k)^{2}=a^{2}$ comparing (1) and (2) we get

$\mathrm{x}-\mathrm{h}=\mathrm{x}-2 \quad \mathrm{y}-\mathrm{k}=\mathrm{y}-2 \quad$ and $\mathrm{a}^{2}=1$

$\mathrm{x} _{1}=1 \cos \theta$ $\mathrm{y} _{1}=1 \sin \theta$

$\therefore$ Required equation of circle is

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

$\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}-4 \mathrm{y}+7=0$

Solution :

Equation of normal at $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ of $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}=0$ is

$\frac{x-x _{1}}{x _{1}-1}=\frac{y-y _{1}}{y _{1}-0}\left(\frac{x-x _{1}}{a x _{1}+g}=\frac{y-y _{1}}{b y _{1}+f}\right)$

Slope of this equation is $\frac{y _{1}}{x _{1}-1}$

Slope of $x+2 y=3$ is $\frac{-1}{2}$

Since given that normal is parallel to $x+2 y=3$

$\therefore \frac{\mathrm{y} _{1}}{\mathrm{x} _{1}-1}=\frac{-1}{2}$

$2 \mathrm{y} _{1}=-\mathrm{x} _{1}+1$ therefore locus of $\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ is $\mathrm{x} _{1}+2 \mathrm{y} _{1}=1$

It is the equation of normal $\Rightarrow(x+2 y)=1$

Solution :

Centre and radius of circle $x^{2}+y^{2}-2 x+4 y+1=0$

is $(1,-2)$ and $\sqrt{1+4-1}=2$ respectively.

If the distance of a line $3 x-4 y=1$ from the centre $(1,-2)$ is equal to radius then the line touches or it is tangent to a circle.

$=\left|\frac{3 \times 1-4(-2)-1}{\sqrt{3^{2}+4^{2}}}\right| \quad\left(\left|\frac{\mathrm{ax} _{1}+\mathrm{by} _{1}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|=\mathrm{d}\right)$

$=\left|\frac{3+8-1}{5}\right|$

$=2$

$\therefore$ line $3 x-4 y=1$ touches the circle.

Let point of contact be $\left(x _{1}, y _{1}\right)$ then equation of tangent to a circle $x^{2}+y^{2}-2 x+4 y+1=0$ is

$\mathrm{xx} _{1}+\mathrm{yy} _{1}-\left(\mathrm{x}+\mathrm{x} _{1}\right)+2\left(\mathrm{y}+\mathrm{y} _{1}\right)+1=0……….(1)$

$\mathrm{x}\left(\mathrm{x} _{1}-1\right)+\mathrm{y}\left(\mathrm{y} _{1}+2\right)-\mathrm{x} _{1}+2 \mathrm{y} _{1}+1=0……..(2)$

and given line $3 x-4 y-1=0$

(1) and (2) are idenfical then comparing (1) and (2) we get

$\frac{\mathrm{x} _{1}-1}{3}=\frac{\mathrm{y} _{1}+2}{-4}=\frac{-\mathrm{x} _{1}+2 \mathrm{y} _{1}+1}{-1}$

$-\mathrm{x} _{1}+1=-3 \mathrm{x} _{1}+6 \mathrm{y} _{1}+3 \quad$ or $\quad 2 \mathrm{x} _{1}-6 \mathrm{y} _{1}-2=0$

$-\mathrm{y} _{1}-2=4 \mathrm{x} _{1}-8 \mathrm{y} _{1}-4 \quad$ or $\quad 4 \mathrm{x} _{1}-7 \mathrm{y} _{1}-2=0$

Solving these two equations of $\mathrm{x} _{1}, \mathrm{y} _{1}$ we get $\mathrm{x} _{1}=\frac{-1}{5}$ and $\mathrm{y} _{1}=\frac{-2}{5}$

$\therefore$ point of contact is $\left(\frac{-1}{5}, \frac{-2}{5}\right)$

Solution :

Let the coordinate of $\mathrm{Pbe}\left(\mathrm{x} _{1}, \mathrm{y} _{1}\right)$ and given circle is $\mathrm{x}^{2}+\mathrm{y}^{2}+4 \mathrm{x}-6 \mathrm{y}+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0$

$(\mathrm{x}+2)^{2}+(\mathrm{y}-3)^{2}-4-9+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0$

$(x+2)^{2}+(y-3)^{2}-9 \sin ^{2} \alpha-13\left(1-\cos ^{2} \alpha\right)=0$

$(x+2)^{2}+(y-3)^{2}-9 \sin ^{2} \alpha-13 \sin ^{2} \alpha=0$

$(x+2)^{2}+(y-3)^{2}=4 \sin ^{2} \alpha=(2 \sin \alpha)^{2}$

$\therefore$ centre and radius of circle is $(-2,3)$ and $2 \sin \alpha$ respectively

Distance of $\mathrm{P}$ and $\mathrm{C}$ is

$\mathrm{PC}=\sqrt{\left(\mathrm{x} _{1}+2\right)^{2}+\left(\mathrm{y} _{1}-3\right)^{2}}$

In ${ } _{\Delta} \mathrm{PCR}$

$\sin \alpha=\frac{2 \sin \alpha}{\sqrt{\left(\mathrm{x} _{1}+2\right)^{2}+\left(\mathrm{y} _{1}-3\right)^{2}}}$

or $\sqrt{\left(x _{1}+2\right)^{2}+\left(y _{1}-3\right)^{2}}=2$

Squaring

$\left(\mathrm{x} _{1}+2\right)^{2}+(\mathrm{y}-3)^{2}=4$ or $(\mathrm{x}+2)^{2}+(\mathrm{y}-3)^{2}=4$

$\therefore$ locus of point $P$ is a circle

Solution :

According to the question

$\sqrt{\mathrm{f}^{2}+\mathrm{g}^{2}-6}=2 \sqrt{\mathrm{f}^{2}+\mathrm{g}^{2}+3 \mathrm{f}+3 \mathrm{~g}}$

Squaring both side

$\mathrm{f}^{2}+\mathrm{g}^{2}-6=4\left(\mathrm{f}^{2}+\mathrm{g}^{2}+3 \mathrm{f}+3 \mathrm{~g}\right)$

$3 \mathrm{f}^{2}+3 \mathrm{~g}^{2}+12 \mathrm{f}+12 \mathrm{~g}+6=0$

Divide by 3 we get $\mathrm{f}^{2}+\mathrm{g}^{2}+4 \mathrm{f}+4 \mathrm{~g}+2=0$

Solution :

Let $\mathrm{P}(\mathrm{a} \cos \theta, \operatorname{asin} \theta)$ be a point on the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2}$ Then equation of chord of contact to the circle $x^{2}+y^{2}=b^{2}$ from $P(a \cos \theta$, asin $\theta)$ is

$x(a \cos \theta)+y(a \sin \theta)=b^{2}$

$\operatorname{axsin} \theta+a y \sin \theta=b^{2}$

It is a tangent to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}$

$\therefore$ length of perpendicular to the line $=$ radius.

$\left|\frac{-\mathrm{b}^{2}}{\sqrt{\mathrm{a}^{2}}}\right|=\mathrm{c}$

$\mathrm{b}^{2}=\mathrm{ac}$

$\therefore$ a, b.c are in G.P.