Solution :

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

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

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

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

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

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

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

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

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

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 $(x_1,y_1)$ 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

$\mathrm{x}-\mathrm{h}=\mathrm{x}-2\mathrm{y}-\mathrm{k}=\mathrm{y}-2$ 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 $({\mathrm{x}}_1,{\mathrm{y}}_1)$ of ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0$ is

$\frac{x-x_1}{x_1-1}=\frac{y-y_1}{y_1-0}(\frac{x-x_1}{a{x}_1+g}=\frac{y-y_1}{b{y}_1+f})$

Slope of this equation is $\frac{y_1}{x_1-1}$

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

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

$\therefore \frac{{\mathrm{y}}_1}{{\mathrm{x}}_1-1}=\frac{-1}{2}$

$2{\mathrm{y}}_1=-{\mathrm{x}}_1+1$ therefore locus of $({\mathrm{x}}_1,{\mathrm{y}}_1)$ is ${\mathrm{x}}_1+2{\mathrm{y}}_1=1$

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

Solution :

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

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

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

$=\left|\frac{3\times 1-4(-2)-1}{\sqrt{3^2+4^2}}\right|(\left|\frac{{\mathrm{ax}}_1+{\mathrm{by}}_1+\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right|=\mathrm{d})$

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

$=2$

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

Let point of contact be $(x_1,y_1)$ then equation of tangent to a circle $x^2+y^2-2x+4y+1=0$ is

${\mathrm{xx}}_1+{\mathrm{yy}}_1-(\mathrm{x}+{\mathrm{x}}_1)+2(\mathrm{y}+{\mathrm{y}}_1)+1=0\dots \dots \dots .(1)$

$\mathrm{x}({\mathrm{x}}_1-1)+\mathrm{y}({\mathrm{y}}_1+2)-{\mathrm{x}}_1+2{\mathrm{y}}_1+1=0\dots \dots ..(2)$

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

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

$\frac{{\mathrm{x}}_1-1}{3}=\frac{{\mathrm{y}}_1+2}{-4}=\frac{-{\mathrm{x}}_1+2{\mathrm{y}}_1+1}{-1}$

$-{\mathrm{x}}_1+1=-3{\mathrm{x}}_1+6{\mathrm{y}}_1+3$ or $2{\mathrm{x}}_1-6{\mathrm{y}}_1-2=0$

$-{\mathrm{y}}_1-2=4{\mathrm{x}}_1-8{\mathrm{y}}_1-4$ or $4{\mathrm{x}}_1-7{\mathrm{y}}_1-2=0$

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 $(\frac{-1}{5},\frac{-2}{5})$

Solution :

Let the coordinate of $\mathrm{Pbe}({\mathrm{x}}_1,{\mathrm{y}}_1)$ and given circle is ${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}+9{\sin }^2\alpha +13{\cos }^2\alpha =0$

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

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

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

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

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

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

$\mathrm{PC}=\sqrt{{({\mathrm{x}}_1+2)}^2+{({\mathrm{y}}_1-3)}^2}$

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

$\sin \alpha =\frac{2\sin \alpha }{\sqrt{{({\mathrm{x}}_1+2)}^2+{({\mathrm{y}}_1-3)}^2}}$

or $\sqrt{{(x_1+2)}^2+{(y_1-3)}^2}=2$

Squaring

${({\mathrm{x}}_1+2)}^2+(\mathrm{y}-3{)}^2=4$ 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

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

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

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

Solution :

Let $\mathrm{P}(\mathrm{a}\cos \theta ,\mathrm{asin}\theta )$ 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

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

$\mathrm{axsin}\theta +ay\sin \theta =b^2$

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.