Question

consider the following equation of a circle.
x^{2}+y^{2}-6x + 6y+9 = 0
step 3 of 3: graph the circle.
answer
press the plot button to add the graph to the grid. points can be moved by dragging or using the arrow keys. points whose positions are related to a moved point will be updated automatically. any curves will update whenever a point is moved.
enable zoom/pan

Explanation:

Step1: Rewrite the equation in standard form

Complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}-6x=(x - 3)^{2}-9$.
For the $y$ - terms: $y^{2}+6y=(y + 3)^{2}-9$.
The original equation $x^{2}+y^{2}-6x + 6y+9 = 0$ becomes $(x - 3)^{2}-9+(y + 3)^{2}-9 + 9=0$.
Simplify it to $(x - 3)^{2}+(y + 3)^{2}=9$.

Step2: Identify the center and radius

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.
For the equation $(x - 3)^{2}+(y + 3)^{2}=9$, the center is $(3,-3)$ and the radius $r = 3$.

Step3: Graph the circle

1. Plot the center point $(3,-3)$ on the coordinate grid.
2. From the center, move 3 units up, down, left, and right to get four points on the circle: $(3,0),(3,-6),(0,-3),(6,-3)$.
3. Sketch the circle passing through these four points.

Answer:

Graph a circle with center $(3,-3)$ and radius 3 on the given grid.