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.
Case (i) :
Consider the rectangular equation of a circle in standard form :
(x - h)^{2} + (y - k)^{2} = r^{2}
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 :
x^{2} + y^{2} + 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)
In each case, convert the given rectangular equation of the circle to parametric form.
Problem 1 :
x^{2} + y^{2} = 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} = r^{2}
Then, we have
(x - 0)^{2} + (y - 0)^{2} = 6^{2}
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} = 7^{2}
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 :
x^{2} + (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 :
x^{2} + y^{2} + 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
x^{2} + y^{2} + 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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