Basic Concepts of Circle

What do we mean by the locus of a point?

Locus of a point is defined to be the path of a point satisfying some geometrical condition; i.e. constraint equations. The path represents a curve, which includes all the points satisfying the given condition.
 

What is a Circle?

A circle is defined as the locus of a point which moves in such a way that its distance from a fixed point is always constant and positive. The fixed point is called the centre of the circle and the given distance the radius of the circle. 

For eg: In real life, when you rotate a stone tied with one end of a string then the path followed by stone is exactly a circle whose centre is your finger an radius is length of the string. 

The equation of a circle with its centre at C(x_c, y_c) and radius r is: 

(x – x_c)² + (y – y_c)² = r² 

Proof: 

Let P(x, y) be any point on the circle. Then by the definition of the locus the constant distance is (see figure given below) 

|PC| = r ⇒ √(x-x_c)² + (y-y_c))²= r 
⇒ (x – x_c)² + (y – y_c)² = r² 

which is the required equation of the circle.

Remark: 
(1) If x_c = y_c = 0 (i.e. the centre of the circle is at origin) then equation of the circle reduce to x² + y² = r². 

(2) If r = 0 then the circle represents a point or a point circle. 
 

Perimeter of the Circle

The perimeter of any figure refers to the sum of the boundaries. Similarly, in case of a circle, the perimeter is given by the circumference of the circle.  

Equation of the circle in Various Forms 

(i) The simplest equations of the circle is x² + y² = r² whose centre is (0, 0) and radius ‘r’. 

(ii) The equation (x – a)²+ (y – b)² = r² represents a circle with centre (a, b) and radius r. 

(iii) The equation x² + y² + 2gx + 2fy + c = 0 is the general equation of a circle with centre (–g, –f) and radius √(g²+f²-c). 

(iv) 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.
 

Equation of a circle under Different Conditions

Parametric Equation of a circle 

Let us consider a circle of radius ‘r’ and centre at C(x_c, y_c) we have: 

(y-y_c)/r = sin θ (see figure given below) 

⇒ y = y_c + r sin θ 

Similarly x = x_c + r cos θ 

This gives the parametric form of the equation of a circle. 
 

Diameter of circle

The locus of middle points of a system of parallel chords of a circle is called the diameter of the circle. The diameter of the circle x² + y² = r² corresponding to the system of parallel chords y = mx + c is x + my = 0.

(a) Every diameter passes through the centre of the circle.

(b) A diameter is perpendicular to the system of parallel chords.
 

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) + R² – a² = 0    …… (1) 

This is the required polar equation.
 

Particular cases of the general equation in polar coordinates

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

r² – 2Rr cos θ + R² – a² = 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 

OP = OA cos θ, 

r = 2a cos θ. 

Remark: 

∠PRQ = π/2 (Angle subtended by diameter at any point on the circle is a right angle). 

⇒ QR ⊥ PR 

⇒ (Slope of QR) x (Slope of PR) = –1 

⇒ (y-y₂)/(x-x₂) ×(y-y₁)/(x-x₁) = – 1 

⇒ (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0 

which gives the required equation.

Note: 

This equation can also be obtained considering 

PR² + QR² = PQ² 
The general from of the equation of a circle is: 
x² + y² + 2gx + 2fy + c = 0        …… (1) 
⇒ (x + g)² + (y + f)² = g² + f² – c 
Comparing this equation with the standard equation (x – x_c)² + (y – y_c)²= r² 

We have: 

Centre of the circle is (–g, –f), Radius = √(g²+f²-c). 
Equation (1) is also written as S = 0.

Remark: 

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
 

Director Circle

The locus of the point of intersection of two perpendicular tangents to a circle is called the director circle. 

We now try to find out the equation of director circle:

If the equation of the circle is x² + y² = a², then the equation of pair of tangents to a circle from a point (x₁, y₁) is 

(x² + y² –a²) (x₁² + y₁² – a²) = (xx₁ + yy₁ – a²)²

If this represents a pair of perpendicular lines then the coefficient of x² + coefficient of y² = 0

i.e. (x₁² + y₁² – a² - x₁²) + (x₁² + y₁² – a² - y₁²) = 0

This gives x₁² + y₁² = 2a²

Hence, this means that the equation of director circle is x² + y² = 2a²

Remark: 

1. Director Circle is a concentric circle whose radius is √2 times the radius of the given circle.

2. The director circle of circle x² + y² + 2gx + 2fy + c = 0 is x² + y² + 2gx + 2fy + 2c - g² - f² = 0.

Illustration: 

Find the centre and the radius 3x² + 3y² – 8x – 10y + 3 = 0. 

Solution: 

We write the given equation as x² + y² – 8/3 x – 10/3 y + 1 = 0. 

⇒ g = -4/3, f = -5/3 , c = 1 

Hence the centre is (4/3,5/3) and the radius is 

√(16/9+25/9-1)=√(32/9)=(4√2)/3.

Illustration: 

Find the length of intercept on y-axis, by a circle whose diameter is the line joining the points (-4, 3) and (12, -1).

Solution: 

The equation of the required circle is (x+4)(x-12) + (y-3)(y+1) = 0

This gives the eqaution as x² + y² – 8x – 2y -51= 0. 

Hence intercept on y-axis is 2√f²-c = 2√(1-(-51)) = 4√13.

Illustration: 

A circle has radius 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 (α, β). It lies on the line y = x – 1 

⇒ β = α – 1. Hence the centre is (α, α – 1). 

⇒ The equation of the circle is (x – α)² + (y – α + 1)² = 9. It passes through (7, 3) 

⇒ (7 – α)² + (4 – α)² = 9 ⇒ 2α² – 22α + 56 = 0 

⇒ α² – 11α + 28 = 0 ⇒ (α – 7) = 0 ⇒ α = 4, 7. 

Hence the required equations are 

x² + y² – 8x – 6y + 6 = 0 and x² + y² – 14x – 12y + 76 = 0.

Illustration: 

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 ± √52. 

Hence the intercept on the y-axis = 22√52 = 4√13.

Illustration: 

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

Solution: 

Let the equation be x² + y² + 2gx + 2fy + c = 0. 

Substituting the coordinates of 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.

Illustration: 

Write general equation of a circle centered at a point on x-axis.

Solution: 

Circle is: x² + y² + 2gx + c = 0, g² – c ≥ 0 
Its centre is (–g, 0) and radius √(g²-c) 
Or 
(x + g)² + (y – 0)² = r² 

Its centre is (–g, 0) and radius r. (figure given above) 

Illustration:  

Write the equation of a circle passing through O (0, 0) A (a, 0) and B (0, b)? Obviously AB is the diameter of the circle. (Figure given below) 

Solution: 

Since the circle passes through O (0, 0) A (a, 0) and B (0, b), hence we have the equation as 

(x – a) (x – 0) + (y – 0) (y – b) = 0

Illustration: 

Find the equation of circle shown in figure given below in polar form.

Solution: 

Clearly, OC is the radius. Hence, we have OP = OA cos θ.

This means r = 2a cos θ, – θ/2 ≤ θ ≤ θ/2, where a is radius of circle.

Illustration: 

Find the co-ordinates of the centre of the circle represented by r = A cos θ + B sin θ. 

Solution: 

r = A cos θ + B sin θ 

= [A/√(A²+B² ) cos θ +B/√(A²+B² ) sin θ ] √(A²+B² ) 

= cos (θ – α) (√(A²+B² )) 

centre is ≡ (1/2 √(A²+B² ),tan_-1 (B/A) )

Note: 

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 is greater than, equal to or less than 0.

S₁ > 0 ⇒ Point is outside the circle

S₁ = 0 ⇒ Point is on the circle

S₁ < 0 ⇒ Point is inside the circle
 

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