# Parametric Equation of a Circle

The equations x = a cosθ , y = a sinθ are called parametric equations of the circle x2 + y2 = a2 and θ is called a parameter.

The point (a cos θ, a sinθ) is also referred to as point θ

The parametric coordinates of any point on the circle

(x – h)2 + (y – k)2 =a2 are given by (h + a cosθ, k + a sinθ) with 0 ≤ θ < 2π

Example : If a straight line through C(-√8, √8), making an angle of 135° with the x-axis, cuts the circle x = 5 cosθ , y = 5 sinθ , in points A and B, find the length of the segment AB.

Solution: The given circle is x2 + y2 = 25 . . .(1)

Equation of line through C is (x + √8)/cos135° = (y − √8)/sin135° = r

Any point on this line is ( − √8 − r/√2 , √8 + r/√2)

If the point lies on the circle x2 + y2 = 25, the

(− √8 − r/√2) + (√8 + r/√2) = 25

⇒ r2 + 8r + 16 − 25 = 0

⇒ r2 + 9r − r − 9 = 0

⇒ r = 1 , − 9

AB = |−9 − 1| = 10.

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

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

Solution:

(i) The radius of the circle is

r2 = (4 − 1)2 + (6 − 2)2

r2 = 25

⇒ r = 5
Hence the equation of the circle is (x − 1)2 + (y − 2)2 = 25

⇒ x2 + y2 − 2x − 4y = 20

(ii) 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 : Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1.

Solution: Let r be the radius of the circle. Then

r = distance of the centre i.e. point (3 , 4) from the line 5x + 12y = 1

= |(15 + 48 − 1)/√(25 + 144)|

= 62/13

Hence the equation of the required circle is (x − 3)2 + (y − 4)2 = (62/13)2

=> x2 + y2 − 6x − 8y + 381/169 = 0

Example : 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

=> x2 + y2 − 8x − 2y − 51 = 0

Intercept made on the y-axis = 2 √(f2 − c)

= 2 √(12 + 51) = 2 √(52)

= 4 √(13)

Example : 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 (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 − 4) (α − 7) = 0

=> α = 4, 7.

Hence the required equations are x2 + y2 − 8x − 6y + 16 = 0 and x2 + y2 − 14x − 12y + 76 = 0.