PARAMETRIC EQUATION OF A CIRCLE

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

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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