Definition
A circle is the locus
of a point which moves in such a way that its distance from a fixed point is
constant. The fixed point is called the centre of the circle and the constant
distance, the radius of the circle.
Equations
of the Circle in Various Forms:
- The simplest equation of the circle
is x2 + y2 = r2 whose centre is (0, 0) and
radius r. - The equation (x – a)2 + (y
– b)2 = r2 represents a circle with centre (a, b) and
radius r. - The equation x2 + y2
+ 2gx + 2fy + c = 0 is the general equation of a circle with centre (-g, -f)
and radius
. - Equation of the circle with points
P(x1, y1) and Q(x2, y2) as
extremities of a diameter is (x – x1)(x – x2) + (y – y1)(y
– y2) = 0.
- The equation of the circle through
three non-collinear points P(x1, y1) , Q(x2, y2)
and R(x3, y3) is
=
0. - Equation
of a circle under different conditions
- CONDITION
-
EQUATION
- (i) Touches both
the axes with centre (a, a)
- and radius a
- (x-a)2 +
(y-a)2 = a2
- (ii) Touches x-axis
only with centre
- (a,
a) and radius a
- (x-a)2
+ (y-a)2 = a2
- (iii)Touches y-axis
only with centre (a, b) and
- radius a
- (x-a)2 +
(y-b)2 = a2
- (iv) Passes through the
origin with centre
-
and radius 
. - x2 +y2
- a x - b x = 0
- .
- Parametric Equation of a Circle
- The equations x = a cosq
, y = a sinq are called parametric equations of
the circle x2+ y2 = a2 and q
is called a parameter. The point (a cos q, a sin q)
is also referred to as point q. The parametric coordinates of any
point on the circle (x – h)2+(y – k)2 = a2 are
given by (h + a cosq, k + a sinq)
with 0 £ q < 2p. .
Example
1
Find the
centre and the radius of the circle 3x2+3y2-8x-10y+3 =
0.
Solution:
We rewrite the given equation as
x2 + y2
- 
x - 
y + 1 = 0
&⇒
g = -
, f = - 
, c = 1
Hence the centre is 
and the radius is
= 
= 
Example
2
Find
the equation of the circle whose centre is (1, 2) and which passes through the
point (4, 6).
Solution:
The radius of the circle is 
= 
= 5.
Hence the equation of
the circle is (x – 1)2 + (y – 2)2 =25
&⇒
x2 + y2 – 2x – 4y = 20
Example
3
Find the equation of a circle passing through (1, 1),
(2, -1) and (3, 2). .
Solution:
Let the equation be
x2 + y2 + 2gx +
2fy + c = 0
Substituting for the three points,
we get
2g + 2f + c = –2
4g – 2f + c = –5
6g + 4f + c = –13
Solving the above three equations, we
obtain:
f = –1/2; g = –5/2 ; c = 4
Hence the equation of the circle is
x2 + y2 – 5x –
y + 4 = 0.
Example
4
Find
the equation of the circle whose diameter is the line joining the points (–4,
3) and (12, –1). Find also the intercept made by it on
the y-axis.
Solution:
The equation of the required circle
is
(x + 4)(x – 12) + (y – 3) (y + 1) = 0
On the y-axis, x = 0
&⇒
-48 + y2 – 2y – 3 = 0
&⇒
y2 – 2y – 51 = 0
&⇒
y = 1 ± 
Hence the intercept on
the y-axis = 
Example
5
A
circle has radius equal to 3 units and its centre lies on the line
y = x – 1. Find the equation of the circle if it passes through (7, 3).
Solution:
Let the centre of the
circle be (a, b). It lies on
the line y = x – 1.
&⇒
b
= a – 1. Hence the centre is (a,
a
– 1).
&⇒
The equation of the circle is
(x – a)2
+ (y – a + 1)2 = 9
It passes through (7,
3)
&⇒
(7 – a)2 + (4 – a)2 =
9
&⇒
2a2
– 22a + 56 = 0
&⇒
a2
– 11a + 28 = 0
&⇒
(a
– 4) (a – 7) = 0
&⇒ a = 4, 7.
Hence the required
equations are
x2 + y2
– 8x – 6y + 16 = 0 and x2 + y2 – 14x – 12y + 76 = 0.
The
position of a point with respect to a circle :
The point P(x1,
y1) lies outside, on, or inside a circle S º
x2 + y2 + 2gx + 2fy + c = 0, according as S1 º
x12 + y12 + 2gx1 + 2fy1
+ c > = or < 0.
