Question

The parametric form of equation of the circle \( x^2 + y^2 - 6x + 2y - 28 = 0 \) is

• A. \( x=-3+\sqrt{38}\cos\theta,\; y=-1+\sqrt{38}\sin\theta \)
• B. \( x=\sqrt{28}\cos\theta,\; y=\sqrt{28}\sin\theta \)
• C. \( x=-3-\sqrt{38}\cos\theta,\; y=-1+\sqrt{38}\sin\theta \)
• D. \( x=3+\sqrt{38}\cos\theta,\; y=-1+\sqrt{38}\sin\theta \)
• E. \( x=3+\sqrt{38}\cos\theta,\; y=-1-\sqrt{38}\sin\theta \)

Hint:

Always complete the square carefully—most mistakes happen in sign handling.

Answer 1

The Correct Option is D

Concept:
• General equation of circle: \[ x^2 + y^2 + Dx + Ey + F = 0 \]
• Convert to standard form by completing squares.
• Parametric form: \[ x = h + r\cos\theta, \quad y = k + r\sin\theta \]

Step 1: Rewriting the equation.
\[ x^2 + y^2 - 6x + 2y - 28 = 0 \] Group terms: \[ (x^2 - 6x) + (y^2 + 2y) = 28 \]

Step 2: Completing the square.
For \(x\): \[ x^2 - 6x = (x-3)^2 - 9 \] For \(y\): \[ y^2 + 2y = (y+1)^2 - 1 \]

Step 3: Substituting back.
\[ (x-3)^2 - 9 + (y+1)^2 - 1 = 28 \] \[ (x-3)^2 + (y+1)^2 = 28 + 9 + 1 \] \[ (x-3)^2 + (y+1)^2 = 38 \]

Step 4: Identifying center and radius.
\[ \text{Center } (h,k) = (3,-1) \] \[ \text{Radius } r = \sqrt{38} \]

Step 5: Writing parametric equations.
\[ x = 3 + \sqrt{38}\cos\theta \] \[ y = -1 + \sqrt{38}\sin\theta \]

Step 6: Final Answer.
\[ \boxed{(4)} \]