Question

Find the parametric equation of the circle $x^2 + y^2 - 6x + 4y - 3 = 0$.

Answer 1

x=3+4\cos t,\quad y=-2+4\sin t, \quad 0\le t\le2\pi.

Answer 2

Step 1. Complete the Square

Given:

$$x^2+y^2-6x+4y-3=0$$

Group x and y terms:

$$(x^2-6x)+(y^2+4y)=3$$

Complete the square for x:

$$x^2-6x=(x^2-6x+9)-9=(x-3)^2-9$$

Complete the square for y:

$$y^2+4y=(y^2+4y+4)-4=(y+2)^2-4$$

Thus,

$$(x-3)^2-9+(y+2)^2-4=3 \quad \Rightarrow \quad (x-3)^2+(y+2)^2=16$$

Step 2. Identify the Center and Radius

Center: $(3,-2)$ Radius: $r=\sqrt{16}=4$

Step 3. Write the Parametric Equations

The standard parametric form of a circle:

$$x = r\cos t + x_0, \quad y = r\sin t + y_0$$

Here, $x_0=3$, $y_0=-2$, and $r=4$. Hence,

$$x = 3+4\cos t, \quad y = -2+4\sin t, \quad 0\le t\le 2\pi.$$

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Minimal Explanation:

Complete the square to get $(x-3)^2+(y+2)^2=16$. The center is $(3,-2)$ and radius is $4$. Thus, the parametric equations are $x=3+4\cos t$ and $y=-2+4\sin t$.