Question

1. what is the standard form of the equation of the circle in the graph?

\\( (x + 3)^2 + (y + 3)^2 = 9 \\)
\\( (x + 3)^2 + (y + 3)^2 = 3 \\)
\\( x^2 + y^2 = 3 \\)
\\( x^2 + y^2 = 9 \\)

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Step2: Identify the center of the circle from the graph

From the graph, the center of the circle is at the origin \((0, 0)\) (since it's on the intersection of the axes, looking at the grid, the center is at \((0,0)\)). Wait, no, wait. Wait, looking at the grid, the center is at \((0,0)\)? Wait, no, the blue circle: let's check the coordinates. Wait, the grid has x and y axes. Wait, maybe I misread. Wait, the center is at \((0,0)\)? Wait, no, the options: let's re - evaluate. Wait, the standard form is \((x - h)^2+(y - k)^2 = r^2\). If the center is \((0,0)\), then \(h = 0\), \(k = 0\). Now, let's find the radius. From the graph, the circle seems to have a radius of 3 (since from the center, it goes 3 units in x and y directions). So \(r = 3\), so \(r^2=9\). So the equation would be \((x - 0)^2+(y - 0)^2=9\), which is \(x^{2}+y^{2}=9\). Wait, but let's check the options. The last option is \(x^{2}+y^{2}=9\). Wait, but wait, maybe I made a mistake in the center. Wait, no, the first two options have center \((- 3,-3)\), but the graph's center is at the origin (0,0). Let's confirm: the blue circle is centered at (0,0) (since the crosshairs are at (0,0) on the grid). The radius: from (0,0) to (3,0) or (0,3) is 3 units (since each grid square is 1 unit? Wait, the x - axis has marks at - 8, - 6, - 4, - 2, 0, 2, 4, 6, 8. So the distance from 0 to 3 (if the radius is 3) would make the radius squared 9. So the equation is \(x^{2}+y^{2}=9\).

Answer:

\(x^{2}+y^{2}=9\) (the last option, which is \(x^{2}+y^{2}=9\))