Maharashtra State BoardHSC Science (Electronics) 11th
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Find the parametric representation of the circle 3x2 + 3y2 − 4x + 6y − 4 = 0 - Mathematics and Statistics

Sum

Find the parametric representation of the circle:

3x2 + 3y2 − 4x + 6y − 4 = 0

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Solution 1

Given equation of the circle is

3x2 + 3y2 − 4x + 6y − 4 = 0

Dividing throughout by 3, we get

`x^2 + y^2 - 4/3x + 2y - 4/3` = 0

∴ `x^2 - 4/3 x + y^2 + 2y - 4/3` = 0

∴ `x^2 - 4/3x + 4/9 - 4/9 + y^2 + 2y + 1 -  1 - 4/3` = 0

∴ `(x^2 - 4/3x + 4/9) + (y^2 + 2y + 1) - 25/9` = 0

∴ `(x - 2/3)^2 + (y + 1)^2 = 25/9`

∴ `(x - 2/3)^2 + [y - (-1)]^2 = (5/3)^2`

Comparing this equation with

(x – h)2 + (y – k)2 = r2, we get

h = `2/3, "k" = -1 and "r" = 5/3`

∴ The parametric representation of the circle in terms of θ are

x = h + r cos θ and y = k + r sin θ

∴ x =`2/3 + 5/3 cos theta and y = -1 + 5/3 sin theta`.

Solution 2

Given equation of the circle is

3x2 + 3y2 − 4x + 6y − 4 = 0

Dividing throughout by 3, we get

`x^2 + y^2 - 4/3x + 2y - 4/3` = 0

∴ `x^2 - 4/3 x + y^2 + 2y - 4/3` = 0

∴ `x^2 - 4/3x + 4/9 - 4/9 + y^2 + 2y + 1 -  1 - 4/3` = 0

∴ `(x^2 - 4/3x + 4/9) + (y^2 + 2y + 1) - 25/9` = 0

∴ `(x - 2/3)^2 + (y + 1)^2 = 25/9`

∴ `(x - 2/3)^2 + [y - (-1)]^2 = (5/3)^2`

Comparing this equation with

(x – h)2 + (y – k)2 = r2, we get

h = `2/3`, k = -1 and `"r" = 5/3`

∴ The parametric representation of the circle in terms of θ are

x = h + r cos θ and y = k + r sin θ

∴ x =`2/3 + 5/3 cos theta and y = -1 + 5/3 sin theta`.

Solution 3

Given equation of the circle is

3x2 + 3y2 − 4x + 6y − 4 = 0

Dividing throughout by 3, we get

`x^2 + y^2 - 4/3x + 2y - 4/3` = 0

∴ `x^2 - 4/3 x + y^2 + 2y - 4/3` = 0

∴ `x^2 - 4/3x + 4/9 - 4/9 + y^2 + 2y + 1 -  1 - 4/3` = 0

∴ `(x^2 - 4/3x + 4/9) + (y^2 + 2y + 1) - 25/9` = 0

∴ `(x - 2/3)^2 + (y + 1)^2 = 25/9`

∴ `(x - 2/3)^2 + [y - (-1)]^2 = (5/3)^2`

Comparing this equation with

(x – h)2 + (y – k)2 = r2, we get

h = `2/3`, k = -1 and `"r" = 5/3`

∴ The parametric representation of the circle in terms of θ are

x = h + r cos θ and y = k + r sin θ

∴ x =`2/3 + 5/3 cos theta and y = -1 + 5/3 sin theta`.

Concept: Parametric Form of a Circle
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