Question

1. graph the circle by rewriting the equation in standard form.

$x^{2}+y^{2}+4x - 4y-1 = 0$

Explanation:

Step1: Group x - terms and y - terms

$$(x^{2}+4x)+(y^{2}-4y)=1$$

Step2: Complete the square for x - terms

For the expression $x^{2}+4x$, we add $(\frac{4}{2})^{2}=4$ to both sides of the equation.
$$(x^{2}+4x + 4)+(y^{2}-4y)=1 + 4$$

Step3: Complete the square for y - terms

For the expression $y^{2}-4y$, we add $(\frac{-4}{2})^{2}=4$ to both sides of the equation.
$$(x^{2}+4x + 4)+(y^{2}-4y+4)=1 + 4+4$$

Step4: Rewrite in standard form

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.
$$(x + 2)^{2}+(y - 2)^{2}=9$$
The center of the circle is $(-2,2)$ and the radius $r = 3$.
To graph:

1. Plot the center point $(-2,2)$ on the coordinate plane.
2. From the center, move 3 units up, down, left and right to get four points on the circle: $(-2,2 + 3)=(-2,5)$, $(-2,2-3)=(-2,-1)$, $(-2 + 3,2)=(1,2)$ and $(-2-3,2)=(-5,2)$.
3. Sketch the circle passing through these points.

Answer:

The standard - form of the circle is $(x + 2)^{2}+(y - 2)^{2}=9$, with center $(-2,2)$ and radius $r = 3$.