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Continue Shopping ```Basic Concepts of Circle

Table of Content

What do we mean by the locus of a point?

What is a Circle?

Perimeter of the Circle

Equation of a circle under Different Conditions

Parametric Equation of a circle

Diameter of circle

General equation of a circle in polar co-ordinate system

Particular cases of the general equation in polar coordinates

Director Circle

The Position of a Point with respect to a Circle

Related Resources

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(xc, yc) and radius r is:

(x – xc)2 + (y – yc)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-xc)2 + (y-yc))2= r
⇒ (x – xc)2 + (y – yc)2 = r2

which is the required equation of the circle.

Remark:
(1) If xc = yc = 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(xc, yc) we have: (y-yc)/r = sin θ (see figure given below)

⇒ y = yc + r sin θ

Similarly x = xc + 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 – xc)2 + (y – yc)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

Related Resources

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