Question

The parametric equations of the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ are:

• A. $x = 2 + 5 \cos \theta, y = 3 + 5 \sin \theta$
• B. $x = -2 + 5 \cos \theta, y = -3 + 5 \sin \theta$
• C. $x = 2 + 25 \cos \theta, y = 3 + 25 \sin \theta$
• D. $x = 5 + 2 \cos \theta, y = 5 + 3 \sin \theta$

Hint:

Center is $(-g, -f)$ and Radius is $\sqrt{g^2+f^2-c}$. For this circle, $g=-2, f=-3, c=-12$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The parametric equations for a circle with center \((h, k)\) and radius \(r\) are \(x = h + r \cos \theta\) and \(y = k + r \sin \theta\).
Step 2: Key Formula or Approach:
For a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\):
Center \(= (-g, -f)\).
Radius \(r = \sqrt{g^2 + f^2 - c}\).
Step 3: Detailed Explanation:
Given circle: \(x^2 + y^2 - 4x - 6y - 12 = 0\).
Comparing coefficients:
\(2g = -4 \implies g = -2\)
\(2f = -6 \implies f = -3\)
\(c = -12\)

Center \((h, k) = (-g, -f) = (2, 3)\).
Radius \(r = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\).

Plugging \(h=2, k=3, r=5\) into the parametric formulas:
\(x = 2 + 5 \cos \theta\)
\(y = 3 + 5 \sin \theta\)
Step 4: Final Answer:
The equations are \(x = 2 + 5 \cos \theta\) and \(y = 3 + 5 \sin \theta\).