Question

what is the radius of this circle?

Explanation:

Step1: Identify the center of the circle

The center of the circle is at the point \((6, 0)\) as seen from the graph (the orange dot).

Step2: Identify a point on the circle

A point on the circle along the x - axis is \((9, 0)\) (or we can also use \((3, 0)\)).

Step3: Calculate the distance between the center and the point on the circle (radius)

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Since \(y_1 = y_2=0\) (both points are on the x - axis), the formula simplifies to \(d=\vert x_2 - x_1\vert\).
Using the center \((6,0)\) and the point \((9,0)\), we have \(r=\vert9 - 6\vert= 3\)? Wait, no, wait. Wait, looking at the graph, the circle goes from \(x = 3\) to \(x=9\) (since the center is at \(x = 6\)). Wait, the distance from \(x = 6\) to \(x=9\) is \(9 - 6=3\)? Wait, no, wait the circle's rightmost point is at \(x = 9\)? Wait, no, the grid lines: the center is at \((6,0)\), and the circle extends from \(x = 3\) to \(x=9\) (since from \(x = 6\), moving 3 units left to \(x = 3\) and 3 units right to \(x = 9\))? Wait, no, wait the x - axis: the center is at \(x=6\), and the circle touches the x - axis at \(x = 3\) and \(x=9\)? Wait, no, looking at the graph, the circle is drawn such that from the center \((6,0)\), the horizontal distance to the edge of the circle is 3 units? Wait, no, wait the circle is from \(x = 3\) to \(x=9\) (since the center is at \(x = 6\), \(9-6 = 3\), \(6 - 3=3\)). Wait, but also, vertically, from \(y=- 3\) to \(y = 3\)? Wait, no, the circle is centered at \((6,0)\) and has a radius. Wait, let's count the grid squares. Each grid square is 1 unit. The center is at \((6,0)\). The rightmost point of the circle is at \(x = 9\) (since from \(x = 6\) to \(x=9\) is 3 units). The leftmost point is at \(x = 3\) (from \(x = 6\) to \(x=3\) is 3 units). The topmost point is at \(y = 3\) and the bottommost at \(y=-3\). So the radius is the distance from the center \((6,0)\) to any point on the circle, say \((9,0)\). Using the distance formula \(r=\sqrt{(9 - 6)^2+(0 - 0)^2}=\sqrt{3^2}=3\)? Wait, no, wait that can't be. Wait, no, wait the circle in the graph: the center is at \((6,0)\), and the circle goes from \(x = 3\) to \(x=9\) (so the diameter is \(9 - 3=6\)), so the radius is half of the diameter, so \(r=\frac{6}{2}=3\)? Wait, no, wait \(9 - 3 = 6\) (diameter), so radius is 3? Wait, but when I look at the graph, the circle seems to have a radius of 3? Wait, no, wait the center is at \((6,0)\), and the circle's right end is at \(x = 9\), so \(9-6 = 3\), so radius is 3. Wait, but maybe I made a mistake. Wait, let's check again. The center is at \((6,0)\). A point on the circle is \((9,0)\). The distance between \((6,0)\) and \((9,0)\) is \(9 - 6=3\). So the radius is 3? Wait, no, wait the circle is drawn with a radius of 3? Wait, but the vertical distance: from \(y = 0\) to \(y = 3\) is 3 units, so the radius is 3.

Wait, no, wait I think I messed up. Wait, the center is at \((6,0)\), and the circle extends from \(x = 3\) to \(x=9\) (so the length from \(x = 3\) to \(x=9\) is 6 units, which is the diameter). So the radius is \(6\div2 = 3\)? Wait, no, \(9-3 = 6\) (diameter), so radius is 3. Yes, that's correct.

Answer:

The radius of the circle is 3.