The parametric equation of the circle `x^(2)+y^(2)+8x-6y=0` are

Options

A. `x=4y+5cos theta, y=3+5 sin theta`
B. `x=-4+5cos theta, y=3+5 sin theta`
C. `x=4+5cos theta, y=-3+5 sin theta`
D. `x=-4+5cos thea,y=-3+5sin theta`

Answer

To find the parametric equations of the circle given by the equation \(x^2 + y^2 + 8x - 6y = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
We start with the equation of the circle:
\[ x^2 + y^2 + 8x - 6y = 0 \]
We will rearrange it to complete the square for both \(x\) and \(y\).

### Step 2: Complete the Square
For \(x\):
\[ x^2 + 8x \quad \text{can be rewritten as} \quad (x + 4)^2 - 16 \]
For \(y\):
\[ y^2 - 6y \quad \text{can be rewritten as} \quad (y - 3)^2 - 9 \]
Now substituting these back into the equation gives:
\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = 0 \]
This simplifies to:
\[ (x + 4)^2 + (y - 3)^2 - 25 = 0 \]
Thus, we have:
\[ (x + 4)^2 + (y - 3)^2 = 25 \]

### Step 3: Identify the Center and Radius
From the equation \((x + 4)^2 + (y - 3)^2 = 25\), we can identify:
- Center: \((-4, 3)\)
- Radius: \(r = \sqrt{25} = 5\)

### Step 4: Write the Parametric Equations
The parametric equations for a circle are given by:
\[ x = h + r \cos \theta \]
\[ y = k + r \sin \theta \]
where \((h, k)\) is the center of the circle.

Substituting the center \((-4, 3)\) and radius \(5\) into the parametric equations:
\[ x = -4 + 5 \cos \theta \]
\[ y = 3 + 5 \sin \theta \]

### Final Parametric Equations
Thus, the parametric equations of the circle are:
\[ x = -4 + 5 \cos \theta \]
\[ y = 3 + 5 \sin \theta \]