Thte parametric form of the equation of circle `4x ^(2) + 4y^(2) =9` is

Options

A. `x=3/2 cos theta, y =3/2 sin theta`
B. `x = 2/3 sin theta, y =2/5 cos theta`
C. `x =3/4 sin theta, y =3/4 cos theta`
D. `x = 3 sin theta, y = sqrt2 cos theta`

Answer

To find the parametric form of the equation of the circle given by \(4x^2 + 4y^2 = 9\), we can follow these steps:

### Step 1: Simplify the Equation
First, we simplify the given equation by dividing both sides by 4:

\[ x^2 + y^2 = \frac{9}{4} \]

### Step 2: Identify the Center and Radius
This equation is now in the standard form of a circle, which is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius. From our equation, we can see that:

- The center \((h, k) = (0, 0)\)
- The radius \(r = \sqrt{\frac{9}{4}} = \frac{3}{2}\)

### Step 3: Write the Parametric Equations
The parametric equations for a circle centered at the origin with radius \(r\) are given by:

\[ x = r \cos(t) \]
\[ y = r \sin(t) \]

Substituting \(r = \frac{3}{2}\):

\[ x = \frac{3}{2} \cos(t) \]
\[ y = \frac{3}{2} \sin(t) \]

### Final Parametric Form
Thus, the parametric form of the equation of the circle \(4x^2 + 4y^2 = 9\) is:

\[ \begin{cases} x = \frac{3}{2} \cos(t) \\ y = \frac{3}{2} \sin(t) \end{cases} \]

where \(t\) varies from \(0\) to \(2\pi\).

---