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General equation of a circle in polar co-ordinate system

Let O be the origin, or pole, OX the initial line, C the centre and ‘a’ the radius of the circle.

Let the polar co-ordinates of C be R and α, so that OC = R and ∠XOC = α . 

Let a radius vector through O at an angle θ with the initial line cut the circle at P and Q. Let OP be r.

Then we have

CP2

= OC2

+ OP2  – 2OC . OP cos COP

i.e. a2

= R2

+ r2  – 2 Rr cos (θ – α) 

i.e. r2  – 2 Rr cos (θ – α) + R

2  – a

2 = 0 …… (1)

This is the required polar equation.

Particular cases of the general equation in polar coordinates.

Note: 

1. Let the initial line be taken to go through the centre C. Then α = 0, and the equation becomes

r2  – 2Rr cos θ + R

2  – a

2= 0.

2. Let the pole O be taken on the circle, so that

R = OC = α

The general equation the becomes

r2  – 2ar cos (θ – α) = 0,

i.e. r = 2a cos (θ – α).

3. Let the pole be on the circle and also let the initial line pass through the centre of the circle. In this case

α = 0, and R = a

Now, the general equation reduces to the simple form r=2a cos θ

This is at once evident from the figure given above.

For, if OCA were a diameter, we have

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Equation (1) is also written as S = 0.

Note:

1. If g2 + f2 – c > 0, circle is real

2. If g2 + f2 – c = 0, circle is a point circle.

3. If g2 + f2 – c < 0, the circle is imaginary.

4. Any second-degree equation ax2 + 2hxy + by2 + 2gx + 2fy+c=0 represents a circle only when h = 0 and a = b i.e. if there is no term

containing xy and co-efficient of x2 and y2 are same, provided abc + 2fgh – af2 – bg2 – ch2 ≠ 0 

1. The equation of the circle through three non-collinear points

2. The circle x2

+ y2

+ 2gx + 2fy + c = 0 makes an intercept on x-axis if x2

+ 2gx + c = 0 has real roots i.e. if g2

> c. And, the magnitude of the

intercept is 2√(g2-c).

The Position of a Point with respect to a Circle

The point P(x1, y1) lies outside, on, or inside a circle S ≡ x2

+ y2

+ 2gx + 2fy + c = 0, according as S1 ≡ x12

+ y12

+ 2gx1 + 2fy1 + c > = or < 0.