The equation of circle is

$(x-h{)}^2+(y-k{)}^2=r^2$

The general equation of circle is

$$ x^2+y^2+2 g x+2 f y+c=0, $$ where $g, f$ and $c$ are constants.

Here, Co-ordinates of the centre are $(-g,-f)$ and

radius $=\sqrt{g^2+f^2-c},\left(g^2+f^2 \geq c\right)$

Let mid-point $M\left(x_1,y_1\right)$ and centre, radius of circle are $(h,k),r$ respectively, then

$\cos \theta =\frac{CM}{r}=\frac{\sqrt{{\left(x_1-h\right)}^2+{\left(y_1-k\right)}^2}}{r}$


$\therefore$ Required locus is $\frac{(x-h{)}^2+(y-k{)}^2-r^2}{r^2}=-{\sin }^2\theta$

The equation of the circle on the line segment joining $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ as diameter is

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0.$

Let the circle $x^2+y^2+2gx+2fy+c=0$

$\begin{array}{rl}\therefore |AB|=\left|x_2-x_1\right|& =\sqrt{{\left(x_2+x_1\right)}^2-4x_1{x}_2}\\ & \\ & =2\sqrt{\left(g^2-c\right)}\end{array}$

$\begin{array}{rl}|CD|=\left|y_2-y_1\right|& =\sqrt{{\left(y_2+y_1\right)}^2-4y_2{y}_1}\\ & \\ & =2\sqrt{\left(f^2-c\right)}\end{array}$

A point $\left(x_1,y_1\right)$ lies outside, on or inside a circle

$S\equiv x^2+y^2+2gx+2fy+c=0$

according as $S_1>$, =, or $<0$

where, $S_1=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c$.

The length of the intercept cut-off from the line $y=m x+c$ by the circle $x^2+y^2=a^2$ is

$$ 2 \sqrt{\left\{\frac{a^2\left(1+m^2\right)-c^2}{\left(1+m^2\right)}\right\}} $$

1. Point form :

The equation of tangent at the point $P\left(x_1,y_1\right)$ to a circle

$x^2+y^2+2gx+2fy+c=0\text{ is }$

$x{x}_1+y{y}_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$

2. Parametric form :

The equation of tangent to the circle $x^2+y^2=a^2$ at the point $(a\cos \theta ,a\sin \theta )$ is

$x\cos \theta +y\sin \theta =a$

3. Slope form :

The equation of a tangent of slope $m$ to the circle $x^2+y^2=a^2$ is

$y=mx\pm a\sqrt{\left(1+m^2\right)}$ and the coordinates of the point of contact are

$\left(\pm \frac{am}{\sqrt{\left(1+m^2\right)}},\mp \frac{a}{\sqrt{\left(1+m^2\right)}}\right)$

1. Point form :

The equation of normal at the point $P\left(x_1,y_1\right)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is

$\frac{x-x_1}{x_1+g}=\frac{y-y_1}{y_1+f}$

2. Parametric form :

Since, parametric coordinates of circle $x^2+y^2=a^2$ is $(a\cos \theta ,a\sin \theta )$.

$\therefore$ Equation of normal at $(a\cos \theta ,a\sin \theta )$ is

$\frac{x}{a\cos \theta }=\frac{y}{a\sin \theta }$

or $\frac{x}{\cos \theta }=\frac{y}{\sin \theta }$ or $y=x\tan \theta$

or $y=mx$, where $m=\tan \theta$

which is slope form of normal.

The length of tangent from the point $P\left(x_1,y_1\right)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is

$\sqrt{\left(x_1^2+y_1^2+2g{x}_1+2f{y}_1+c\right)}=\sqrt{S_1}$

The length of tangent from the point $P\left(x_1,y_1\right)$ to the circle $x^2+y^2+2gx+2fy+c=0$ is

$\sqrt{\left(x_1^2+y_1^2+2g{x}_1+2f{y}_1+c\right)}=\sqrt{S_1}$

The equation of the chord of contact of tangents drawn from a point $\left(x_1,y_1\right)$ to the

circle $x^2+y^2=a^2$ is $x{x}_1+y{y}_1=a^2$

The equation of the chord of the circle $x^2+y^2=a^2$ bisected at the point $\left(x_1,y_1\right)$ is given by

$x{x}_1+y{y}_1-a^2=x_1^2+y_1^2-a^2$

or $T=S_1$

The combined equation of the pair of tangents drawn from a point $P\left(x_1, y_1\right)$ to the circle $x^2+y^2=a^2$ is

$$ \left(x^2+y^2-a^2\right)\left(x_1^2+y_1^2-a^2\right)=\left(x x_1+y y_1-a^2\right)^2 $$

or $ S S_1=T^2$

where, $ S=x^2+y^2-a^2, S_1=x_1^2+y_1^2-a^2$

and $ T=x x_1+y y_1-a^2$

The equation of the director circle of the circle $x^2+y^2=a^2$ is

$x^2+y^2=2a^2$

Then, distance between their centres $=$ sum of their radii i.e.

$\left|C_1 C_2\right|=r_1+r_2$

Then, distance between their centres $=$ Difference of their radii

$\text{ i.e. }\left|C_1{C}_2\right|=\left|r_1-r_2\right|$

The equation of common chord of two circles

$$ S \equiv x^2+y^2+2 g x+2 f y+c=0 $$

and $ S^{\prime} \equiv x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}=0$ is

$ 2 x\left(g-g^{\prime}\right)+2 y\left(f-f^{\prime}\right)+c-c^{\prime}=0$ i.e.

$ S-S^{\prime}=0$



Length of common chord :

We have, $\quad P Q=2(P M) \quad(\because M$ is mid-point of $P Q)$

$$ =2 \sqrt{\left\{\left(C_1 P\right)^2-\left(C_1 M\right)^2\right\}} $$

where, $C_1 P=$ radius of the circle $(S=0)$

and $C_1 M=$ length of perpendicular from $C_1$ on common chord $P Q$.

Let the circle be $x^2+y^2+2gx+2fy+c=0$ and line mirror is $lx+my+n=0$ in this condition, radius of circle remains unchanged but centres changes. Let the centre of imaged circle be $\left(x_1,y_1\right)$.

Then, $\frac{x_1-(-g)}{l}=\frac{y_1-(-f)}{m}=\frac{-2(-lg-mf+c)}{\left(l^2+m^2\right)}$

we get, $x_1=\frac{\left(l^{2}g-m^{2}g+2mlf-2nl\right)}{\left(l^2+m^2\right)}$

and $y_1=\frac{\left(m^{2}f-l^{2}f+2mlg-2mn\right)}{\left(l^2+m^2\right)}$