Graph x^2+y^2=2^2

The given equation is that of a circle in the Cartesian coordinate system. The general form of a circle's equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle, and r is the radius. The question appears to be asking for the graph of a circle centered at the origin (0,0) with a radius of 2 units. You'd be expected to plot this circle on a coordinate plane, showing all the points (x,y) that satisfy the equation x^2 + y^2 = 2^2.

$x^{2} + y^{2} = 2^{2}$

Answer

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Solution:

Step 1:

Compute the square of $2$ . The equation becomes $x^{2} + y^{2} = 2^{2}$ which simplifies to $x^{2} + y^{2} = 4$ .

Step 2:

Recognize that the equation is in the standard form of a circle's equation. Identify the center and radius using the general form $(x - h)^{2} + (y - k)^{2} = r^{2}$ .

Step 3:

Compare the given equation with the standard form to find the values of $r$ , $h$ , and $k$ . Here, $r$ denotes the circle's radius, $h$ is the horizontal shift from the origin, and $k$ is the vertical shift from the origin. We find that $r = 2$ , $h = 0$ , and $k = 0$ .

Step 4:

Determine the circle's center using the coordinates $(h, k)$ . The center is at $(0, 0)$ .

Step 5:

Summarize the key information needed to graph the circle. The center is at $(0, 0)$ and the radius is $2$ .

Step 6:

Graph the circle using the identified center and radius.

Knowledge Notes:

To graph an equation of the form $x^{2} + y^{2} = r^{2}$ , we must understand that this represents a circle in Cartesian coordinates with a center at the origin $(0, 0)$ and a radius $r$ . The general form of a circle's equation is $(x - h)^{2} + (y - k)^{2} = r^{2}$ , where $(h, k)$ is the center of the circle and $r$ is the radius. The steps to graphing a circle include:

Simplifying the given equation if necessary.
Identifying the center $(h, k)$ by comparing the given equation to the general form.
Determining the radius $r$ .
Plotting the center on the coordinate plane.
Drawing the circle with the identified radius, ensuring that all points on the circle are equidistant from the center.

In this problem, since there are no $h$ and $k$ values in the equation, it implies that the center of the circle is at the origin. The radius is the square root of the constant term on the right side of the equation. After plotting the center, use a compass or a round object to trace out the circle with the appropriate radius on a graph.