Find the parametric equation for the equation

Question

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

 

Answer 1

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.

 

 

Comments

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.