Posted on - 15-01-2017

IIT JEE

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.

- The simplest equation of the circle
is x
^{2}+ y^{2}= r^{2}whose centre is (0, 0) and

radius r. - The equation (x – a)
^{2}+ (y – b)^{2}= r^{2}represents a circle with centre (a, b) and

radius r. - The equation x
^{2}+ y^{2}+ 2gx + 2fy + c = 0 is the general equation of a circle with centre (-g, -f) and radius . - Equation of the circle with points
P(x
_{1}, y_{1}) and Q(x_{2}, y_{2}) as extremities of a diameter is (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0. - The equation of the circle through
three non-collinear points P(x
_{1}, y_{1}) , Q(x_{2}, y_{2}) and R(x_{3}, y_{3}) 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}= a^{2} - (ii) Touches x-axis only with centre
- (a, a) and radius a
- (x-a)
^{2}+ (y-a)^{2}= a^{2} - (iii)Touches y-axis only with centre (a, b) and
- radius a
- (x-a)
^{2}+ (y-b)^{2}= a^{2} - (iv) Passes through the origin with centre
- and radius .
- x
^{2}+y^{2}- 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 x
^{2}+ y^{2 }= a^{2}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}= a^{2}are given by (h + a cosq, k + a sinq) with 0 £ q < 2p. .

Find the
centre and the radius of the circle 3x^{2}+3y^{2}-8x-10y+3 =
0.

We rewrite the given equation as

x^{2} + y^{2}
- x - y + 1 = 0

&⇒
g = -, f = - , c = 1

Hence the centre is and the radius is

= =

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

The radius of the circle is = = 5.

Hence the equation of
the circle is (x – 1)^{2} + (y – 2)^{2} =25

&⇒
x^{2} + y^{2} – 2x – 4y = 20

Find the equation of a circle passing through (1, 1), (2, -1) and (3, 2). .

Let the equation be

x^{2} + y^{2} + 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

x^{2} + y^{2} – 5x –
y + 4 = 0.

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.

The equation of the required circle is

(x + 4)(x – 12) + (y – 3) (y + 1) = 0

On the y-axis, x = 0

&⇒
-48 + y^{2} – 2y – 3 = 0

&⇒
y^{2} – 2y – 51 = 0

&⇒
y = 1 ±

Hence the intercept on the y-axis =

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).

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

&⇒
2a^{2}
– 22a + 56 = 0

&⇒
a^{2}
– 11a + 28 = 0

&⇒
(a
– 4) (a – 7) = 0

&⇒ a = 4, 7.

Hence the required equations are

x^{2} + y^{2}
– 8x – 6y + 16 = 0 and x^{2} + y^{2} – 14x – 12y + 76 = 0.

The point P(x_{1},
y_{1}) lies outside, on, or inside a circle S º
x^{2} + y^{2} + 2gx + 2fy + c = 0, according as S_{1} º
x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1}
+ c > = or < 0.