- A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.
- The fixed point is called the centre of the circle
- The distance from the centre to a point on the circle is called the radius of the circle
- The general equation of a circle with center \( (h, k) \) and radius \( r \) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \] - The special case where the center is the origin

\( x^2 + y^2 = r^2 \).

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

Center is given by (-g, -h)

Radius of the Circle

$r=\sqrt {g^2 + f^2 -c}$

Let A(h,k) be any point on the circle

Equation of line AP will be

$y -y_1= \frac {k-y_1}{h-x_1} (x- x_1)$

Equation of line AQ will be

$y -y_2= \frac {k-y_2}{h-x_2} (x- x_2)$

Since two lines are Perpendicular in semi circle

$\frac {k-y_1}{h-x_1}. \frac {k-y_2}{h-x_2}=-1$ $(h-x_1)(h-x_2)+ (k-y_1)(k-y_2)=0$

So for any point x, y we can write as

$(x-x_1)(x-x_2)+ (y-y_1)(y-y_2)=0$

A point $(x_1,y_1)$ lies outside the circle if $S_1 ≡ x_1^2+y_1^2+2gx_1+2fy_1+c$ is positive ($S_1 > 0$).

A point $(x_1,y_1)$ lies inside the circle if $S_1 ≡ x_1^2+y_1^2+2gx_1+2fy_1+c$ is negative ($S_1 < 0$).

A point $(x_1,y_1)$ lies one the circle if $S_1 ≡ x_1^2+y_1^2+2gx_1+2fy_1+c$ is Zero ($S_1 = 0$).

p=r, => Line touches the circle.

p>r, => Line is outside the circle.

p

p=0, => Line is the diameter of the circle.

Equation of tangent at $P(x_1,y_1)$ in a circle

For Circle as center as Origin

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

Equation of tangent

$xx_1 + yy_1=a^2$

For general Circle

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

Equation of tangent is

$xx_1 + yy_2 + g(x+x_1) + f(y+y_1) + c=0$

The circle $x^2+y^2 = a^2$ touches the line $y = mx+c$ if $ c = \pm a \sqrt {(1+m^2)}$

(a) If $a^2(1+m^2) – c^2 > 0$, the line will meet the circle at real and different points.

(b) If $c^2 = a^2(1+m^2)$, the line will touch the circle.

(c) If $a^2(1+m^2) – c^2 <0$, the line will meet the circle at two imaginary points.

From this condition of tangency

Line $y = mx \pm a \sqrt {(1+m^2)}$ is a tangent of the circle $x^2+y^2 =a^2$

Normal to any point on the circle is the line which passes through point and its perpendicular to the tangent at that point

For Circle as center as Origin

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

Equation of Normal

$xy_1 - yx_1=0$

For general Circle

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

Equation of Normal is

$y-y_1 = \frac {y_1+f}{x1+g}(x-x_1)$

Two tangents can be drawn from a point $P(x_1, y_1)$ to a circle. Let PQ and PR be two tangents drawn from P(x1, y1) to the circle x2+y2+2gx+2fy+c = 0. Then, PQ = PR is called the length of the tangent drawn from the point and is given by

$PQ = PR = \sqrt {(x_1^2+y_1^2+2gx_1+2fy_1+c)}$

For Circle as center as Origin

$x^2 + y^2 =a^2$ $PQ = PR = \sqrt {(x_1^2+y_1^2-a^2)}$

Two tangents, PQ and PR, can be drawn from a point $P(x_1, y_1)$ to the circle $S = x^2+y^2+2gx+2fy+c = 0$.

The combined equation is given by

$(x^2+y^2+2gx+2fy+c)(x_1^2+y_1^2+2gx_1+2fy_1+c)= (xx_1 + yy_2 + g(x+x_1) + f(y+y_1) + c)^2$

This can be written like below to remember easily

$SS_1 = T^2$

Where S = 0 is the equation of the circle, T = 0 is the equation of tangent at $(x_1, y_1)$, and S1 is obtained by replacing x by $x_1$ and y by $y_1$ in S.

The chord joining the two points of contact of tangents to a circle drawn from any point P is called the chord of contact of P with respect to the given circle.

For Circle as center as Origin

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

Equation of Chord of Contact

$xx_1 + yy_1=a^2$

For general Circle

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

Equation of Chord of Contact

$xx_1 + yy_2 + g(x+x_1) + f(y+y_1) + c=0$

If the point (x1, y1) lies on the circle, then the chord of contact coincides with the equation of the tangent.

$y-y_1 = \frac {-x_1}{y_1}(x-x_1)$ or $xx_1+yy_1 = x_1^2+y_1^2$, which can be represented by $T = S_1$

Let $S_1$ and $S_2$ be the 2 circles

$S_1 = x^2+y^2+2g_1x+2f_1y+c_1 = 0$ and $S_2= x^2+y^2+2g_2x+2f_2y+c_2 = 0$

Then, the equation of the common chord is $S_1-S_2 = 0$

$2x(g_1-g_2) + 2y(f_1-f_2) + c_1-c_2 = 0$

The required equation for the family of circles passing through the point of intersection of circle S = 0 and line L = 0 is given by $S+ \lambda L = 0$, where $\lambda$ is a parameter.

The equation of the family of circles passing through the point of intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$. Where $\lambda\ne -1$.

Note that both circles should be in the given format of $S_1$ and $S_2$.

If $r_1$ and $r_2$ are the radius of the circles and d is the distance between the center, then circles will be orthogonal if

$d^2 = r_1^2 + r_2^2$

This is because the radii of the circles at the points of intersection will form a right angle with the line connecting the centers.

For the General equation of Circles

$S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$

$S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$

$S_1$ and $S_2$ are orthogonal if $2g_1g_2 + 2f_1f_2 = c_1+ c_2$

If the sum of the radii of two circles is equal to the distance between the centres, then the circles touch externally. The circles will satisfy the condition

$c_1c_2 = r_1 + r_2$

If the difference between the radii and the distance between the centres are equal, then the circles touch internally. The circles will satisfy the condition

$c_1c_2 = r_1 – r_2$