Circles
1.0 Definition
2.0 Equation of circle in various forms
2.1 Central Form
2.2 Standard Form
2.3 Parametric Form
2.4 General form of circle
2.4.1 Application of general form of circle
2.5 Diametric form of circle
3.0 Intercepts made by a circle on coordinate axis
4.0 Position of a point with respect to a circle
5.0 Maximum and minimum distance of a point from a circle
6.0 Intersection of a line and a circle
7.0 Length of intercept cutoff from a line by a circle or length of chord of a circle
8.0 Equation of tangent to a circle
8.1 Slope form
8.2 Point form
8.3 If a point outside the circle is given through which tangent to a circle passes, then
8.4 Parametric form
9.0 Tangents from a point to the circle
10.0 Length of tangent from a point to a circle
11.0 Common Tangents
12.0 Equation of common tangents
13.0 Pair of tangents
14.0 Normal to a circle at a given point
15.0 Common chord of two circles
16.0 Equation of chord joining two points on circle
17.0 Equation of chord of circle whose midpoint is given
18.0 Chord of contact
19.0 Orthogonal Circles
20.0 Director Circle
21.0 Family of circles
2.4 General form of circle
2.2 Standard Form
2.3 Parametric Form
2.4 General form of circle
2.4.1 Application of general form of circle
2.5 Diametric form of circle
8.2 Point form
8.3 If a point outside the circle is given through which tangent to a circle passes, then
8.4 Parametric form
As we know that, central form of equation of a circle is $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {a^2}$$ where centre of circle is $(h,k)$ and radius of circle is $a$. 
$${x^2} + {y^2} - 2hx - 2ky + {h^2} + {k^2} - {a^2} = 0...(1)$$ which is of the form $${x^2} + {y^2} + 2gx + 2fy + c = 0$$ On comparing equation $(1)$ and $(2)$, we get $h = - g,{\text{ }}k = - f,{\text{ }}and{\text{ }}a = \sqrt {{g^2} + {f^2} - c} $
Therefore, coordinates of centre is $(-g,-f)$ and radius $\sqrt {{g^2} + {f^2} - c} $ such that ${g^2} + {f^2} \geqslant c$ where $g$, $f$ and $c$ are constants.
How to find the centre and radius of a circle if general form of circle is given:
In general form, make coefficients of ${x^2}$ and ${y^2}$ equal to $1$ and right hand side of the equation equal to $0$. Coordinates of centre be $(\alpha ,\beta )$ such that $$\alpha = \frac{{ - 1}}{2} \times coefficient{\text{ }}of{\text{ }}x$$ and $$\beta = \frac{{ - 1}}{2} \times coefficient{\text{ }}of{\text{ }}y$$ and $$radius=\sqrt {{\alpha ^2} + {\beta ^2} - constant{\text{}}\ term} $$
Question 4. The equation of circle is $5{x^2} - 5{y^2} - 4x - 3y = 1$. Find the coordinates of centre of circle and its radius.
Solution: We can write the equation of circle as $${x^2} - {y^2} - \frac{4}{5}x - \frac{3}{5}y - \frac{1}{5} = 0$$
Therefore, the coordinates of centre of circle is $\left( {\frac{2}{5},\frac{3}{{10}}} \right)$.
$\alpha = \frac{{ - 1}}{2} \times coefficient{\text{ }}of{\text{ }}x = - \frac{1}{2} \times - \frac{4}{5} = \frac{2}{5}$ and
$\beta = \frac{{ - 1}}{2} \times coefficient{\text{ }}of{\text{ }}y= - \frac{1}{2} \times - \frac{3}{5} = \frac{3}{{10}}$
Radius$= \sqrt {{\alpha ^2} + {\beta ^2} - constant\ {\text{}}term}=\sqrt {\frac{{{2^2}}}{5} + \frac{{{3^2}}}{{10}} - \frac{1}{5}} = \sqrt {\frac{4}{{25}} + \frac{9}{{10}} - \frac{1}{5}} = \sqrt {\frac{5}{{10}}} = \frac{1}{{2\sqrt 5 }}$.
