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
\( x^2 + y^2 = r^2 \).
General Equation of Circle
The General equation of Circle is
$x^2 + y^2 + 2gx + 2fy + c=0$
Center is given by (-g, -h)
Radius of the Circle
$r=\sqrt {g^2 + f^2 -c}$
Equation of Circle when points of extremities of diameter are Given
Here we have to find Equation of Circle with points $P(x_1, y_1)$ and $Q(x_2, y_2)$ as extremities of diameter
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
$(x-x_1)(x-x_2)+ (y-y_1)(y-y_2)=0$
Position of a Point with Respect to a Circle
Let the circle Equation $S ≡ x^2+y^2+2gx+2fy+c = 0$
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 ($S_1 = 0$).
Line and a Circle
Let L = 0 be a line, and S = 0 be a circle; if r is the radius of a circle and p is the length of the perpendicular from the centre of the circle to the line, then if
p=r, => Line touches the circle.
p>r, => Line is outside the circle.
p Line is the chord of the circle.
p=0, => Line is the diameter of the circle.
Equation of Tangent to the circle
Equation of tangent at $P(x_1,y_1)$ in a circle Case A
For Circle as center as Origin
$x^2 + y^2 =a^2$
Equation of tangent
$xx_1 + yy_1=a^2$
Case B
For general Circle
$x^2 + y^2 + 2gx + 2fy + c=0$
Equation of tangent is
$xx_1 + yy_2 + g(x+x_1) + f(y+y_1) + c=0$
Condition for Tangency
A line L = 0 touches the circle S = 0, if the length of the perpendicular drawn from the centre of the circle to the line is equal to the radius of the circle, i.e., p = r. This is the condition of tangency for the line L = 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 $x^2+y^2 =a^2$
Equation of Normal to Circle
Normal to any point on the circle is the line which passes through point and its perpendicular to the tangent at that point Case A
For Circle as center as Origin
$x^2 + y^2 =a^2$
Equation of Normal
$xy_1 - yx_1=0$
Case B
For general Circle
$x^2 + y^2 + 2gx + 2fy + c=0$
Equation of Normal is
$y-y_1 = \frac {y_1+f}{x1+g}(x-x_1)$
Length of Tangent of Circle
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
$x^2 + y^2 =a^2$
$PQ = PR = \sqrt {(x_1^2+y_1^2-a^2)}$
Pair of Tangents to Circle
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.
Chord of Contact
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. Case A
For Circle as center as Origin
$x^2 + y^2 =a^2$
Equation of Chord of Contact
$xx_1 + yy_1=a^2$ Case B
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.
Equation of chord whose midpoint is given
The equation of the chord of the circle $x^2+y^2 = a^2$ whose midpoint $P(x_1, y_1)$ given is
$y-y_1 = \frac {-x_1}{y_1}(x-x_1)$ or $xx_1+yy_1 = x_1^2+y_1^2$, which can be represented by $T = S_1$
Common chord of two circles
The line joining the points of intersection of two circles is called the common chord.
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 $S_1-S_2 = 0$
$2x(g_1-g_2) + 2y(f_1-f_2) + c_1-c_2 = 0$
Family of Circles
Family of circles passing through the point of intersection of a line and circle
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.
Family of circles passing through the point of intersection of two circles
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 $S_2$.
Orthogonal Circles
"Orthogonal circles" refers to two circles that intersect at a right angle. This means that the tangent lines to each circle at the points of intersection are perpendicular to each other.
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 $2g_1g_2 + 2f_1f_2 = c_1+ c_2$
Contact of Circles
Case 1
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
$c_1c_2 = r_1 + r_2$
Case 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
$c_1c_2 = r_1 – r_2$
