PARAMETRIC EQUATION OF A CIRCLE

The following is the equation of a circle in parametric form :

x = h + r cosθ

y = k + r sinθ

where θ is the parameter, (h, k) is the centre and r is the radius of the circle.

Converting Rectangular Form to Parametric Form

Case (i) :

Consider the rectangular equation of a circle in standard form :

(x - h)2 + (y - k)2 = r2

When the rectangular equation of a circle is given in standard form, we can find the center (h, k) and radius r of the circle easily. Once we get the center (h, k) and radius r, we can substitute the values for h, k and r into the equation of the circle in parametric form given above.

Case (ii) :

Consider the rectangular equation of a circle in general form :

x2 + y2 + 2gx + 2fy + c = 0

When the rectangular equation of a circle is given in general form. We can use the formulas given below to find the center (h, k) and radius r.

Center (h, k) = (-g, -f)

Video Lesson

Solved Problems

In each case, convert the given rectangular equation of the circle to parametric form.

Problem 1 :

x2 + y2 = 36

Solution :

The given rectangular equation of the circle is in standard form. It can be written in the exact form 

(x - h)2 + (y - k)2 = r2

Then, we have

(x - 0)2 + (y - 0)2 = 62

In the above rectangular equation of the circle,

center (h, k) = (0, 0)

radius r = 6

Equation of a circle in parametric form :

x = h + r cosθ

y = k + r sinθ

Substitute h = 0, k = 0 and r = 6.

x = 0 + 6 cosθ ----> x = 6 cosθ

y = 0 + 6 sinθ ----> y = 6 sinθ

Problem 2 :

(x + 2)2 + (y - 3)2 = 49

Solution :

The given rectangular equation is in standard form and it can be written as 

[x - (-2)]2 + (y - 3)2 = 72

In the above rectangular equation of the circle,

center (h, k) = (-2, 3)

radius r = 7

Equation of a circle in parametric form :

x = h + r cosθ

y = k + r sinθ

Substitute h = -2, k = 3 and r = 7.

x = -2 + 7 cosθ

y = 3 + 7 sinθ

Problem 3 :

x2 + (y + 7)2 = 10

Solution :

The given rectangular equation is in standard form and it can be written as 

(x - 0)2 + [y - (-7)]2 = (√10)2

In the above rectangular equation of the circle,

center (h, k) = (0, -7)

radius r = √10

Equation of a circle in parametric form :

x = h + r cosθ

y = k + r sinθ

Substitute h = 0, k = -7 and r = √10.

x = √10 cosθ

y = -7 + √10 sinθ

Problem 4 :

x2 + y2 + 8x - 10y + 5 = 0

Solution :

The given rectangular equation is in general form. So, we have to find the center and radius.

Comparing the given equation of the circle with

x2 + y2 + 2gx + 2fy + c = 0,

we get

2g = 8

g = 4

-g = -4

2f = 10

f = -5

-f = 5

c = 5

Center :

(h, k) = (-g, -f)

= (-4, 5)

Radius :

Equation of a circle in parametric form :

x = h + r cosθ

y = k + r sinθ

Substitute h = -4, k = 5 and r = 6.

x = -4 + 6 cosθ

y = 5 + 6 sinθ

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