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Find the parametric equation for the equation

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x^2 + y^2 + 12x + 4y - 3 = 0

 

asked Jul 5, 2013 in PRECALCULUS by harvy0496 Apprentice

1 Answer

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The given parametric equation is x^2+y^2+12x+4y -3 = 0

Add 36 to each side.

x^2+12x+36+y^2+4y = 3+36

(x+6)^2+y^2+4y = 39

Add 4 to each side

(x+6)^2+y^2+4y+4 = 39+4

(x+6)^2+(y+2)2 = 43

(x -(-6))^2+(y-(-2))^2 = (√43)^2

The equation is compare to circle equation

center is (h, k) = ( -6, -2) and radius(r) = √43.

 

 

answered Jul 5, 2013 by goushi Pupil

The rectangular equation of circle : (x - h)2 + (y - k)2 = r2.

The parametric equation of circle : x = h + r cos(θ) and y = k + r sin(θ), where θ is parameter constant.

 

The rectangular equation of circle : [x - (-6)]2 + [y - (-2)]2 = (√43)2

The parametric equation of circle : x = - 6 + √43cos(θ) and y = - 2 + √43 sin(θ), where θ is parameter constant.

 

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