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. None of these

Hint:

For $x^2+y^2+2gx+2fy+c=0$, the center is $(-g, -f)$ and $r = \sqrt{g^2+f^2-c}$.

Answer 1

The Correct Option is A

Step 1: Concept
To find the parametric equations, we first convert the general equation of the circle into the standard form $(x-h)^2 + (y-k)^2 = r^2$.

Step 2: Meaning
Complete the square for $x$ and $y$: $(x^2 - 4x + 4) + (y^2 - 6y + 9) = 12 + 4 + 9$, which simplifies to $(x-2)^2 + (y-3)^2 = 25$.

Step 3: Analysis
From the standard form, the center $(h, k)$ is $(2, 3)$ and the radius $r$ is $\sqrt{25} = 5$.

Step 4: Conclusion
The parametric equations are $x = h + r \cos \theta$ and $y = k + r \sin \theta$. Substituting our values gives $x = 2 + 5 \cos \theta$ and $y = 3 + 5 \sin \theta$. Final Answer: (A)