Question

If \( \theta \) is a parameter, then the parametric equations of the circle \[ x^2 + y^2 - 6x + 4y - 3 = 0 \] are given by

• A. \( x = -3 + 4 \cos \theta \) and \( y = -2 + 4 \cos \theta \)
• B. \( x = 3 + 4 \cos \theta \) and \( y = -2 + 4 \sin \theta \)
• C. \( x = 3 + 4 \sin \theta \) and \( y = 2 + 4 \cos \theta \)
• D. \( x = 3 + 4 \cos \theta \) and \( y = 2 + 4 \sin \theta \)

Hint:

To find the parametric equations of a circle, first complete the square to convert the equation into standard form, then use the formula \( x = h + r \cos \theta \) and \( y = k + r \sin \theta \).

Answer 1

The Correct Option is D

Step 1: Completing the square.
To convert the given equation of the circle to a standard form, we complete the square for both \( x \) and \( y \). The equation is: \[ x^2 - 6x + y^2 + 4y = 3. \] Completing the square for \( x \) and \( y \), we get: \[ (x - 3)^2 + (y + 2)^2 = 16. \]
Step 2: Standard form of the circle.
The equation is now in the standard form of a circle: \[ (x - 3)^2 + (y + 2)^2 = 4^2, \] which represents a circle with center \( (3, -2) \) and radius 4.
Step 3: Parametric equations.
The parametric equations for a circle with center \( (h, k) \) and radius \( r \) are: \[ x = h + r \cos \theta, \quad y = k + r \sin \theta. \] Substituting \( h = 3 \), \( k = -2 \), and \( r = 4 \), we get: \[ x = 3 + 4 \cos \theta, \quad y = -2 + 4 \sin \theta. \]
Step 4: Conclusion.
Thus, the correct parametric equations are given by option (D).