Question

determine the equation of the circle graphed below.

Explanation:

Step1: Recall the standard circle equation

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

Step2: Find the center of the circle

From the graph, the center of the circle is at \((5, 0)\) (since it's on the x - axis at \(x = 5\) and \(y=0\)).

Step3: Determine the radius

The radius is the distance from the center to any point on the circle. Looking at the graph, the distance from the center \((5,0)\) to the top of the circle (e.g., \((5, 3)\) or calculating the vertical distance from \(y = 0\) to the top of the circle which is \(y=3\) (since the circle goes from \(y = 0\) to \(y = 6\)? Wait, no, let's check the grid. The center is at \((5,0)\), and the circle extends up to \(y = 3\)? Wait, no, looking at the graph, the circle's top is at \(y = 3\)? Wait, no, the circle is drawn such that the center is \((5,0)\) and the radius is 3? Wait, no, let's count the grid squares. From the center \((5,0)\) to the rightmost point: the rightmost point is at \(x=8\), so the distance from \(x = 5\) to \(x = 8\) is \(8 - 5=3\). So the radius \(r = 3\).

Step4: Substitute \(h\), \(k\), and \(r\) into the equation

Substitute \(h = 5\), \(k = 0\), and \(r = 3\) into \((x - h)^2+(y - k)^2=r^2\). We get \((x - 5)^2+(y - 0)^2=3^2\), which simplifies to \((x - 5)^2+y^2 = 9\). Wait, wait, no, wait the top of the circle: from \(y = 0\) (center) to the top of the circle, let's see the y - coordinate. The circle's top is at \(y = 3\)? Wait, no, looking at the graph, the circle is from \(y = 0\) (bottom) up to \(y = 6\)? Wait, no, the center is at \((5,0)\), and the circle touches the x - axis at the center? No, wait the circle is tangent to the x - axis at the center? Wait, no, the bottom of the circle is on the x - axis at \((5,0)\)? Wait, no, the circle is drawn such that the bottom point is \((5,0)\), and the center is \((5,3)\)? Wait, I made a mistake. Let's re - examine the graph. The circle is above the x - axis? Wait, no, the circle touches the x - axis at a point. Wait, the center: let's find the center. The circle is symmetric about the vertical line \(x = 5\) (since the bottom points are from \(x = 3\) to \(x = 7\) on the x - axis? Wait, no, the circle intersects the x - axis at \((3,0)\) to \((7,0)\)? Wait, no, the circle is tangent to the x - axis? Wait, no, the bottom of the circle is on the x - axis, and the center is at \((5,3)\) because the radius is the distance from the center to the x - axis (which is the y - coordinate of the center). Let's count the grid: from the x - axis (y = 0) up to the center, if the radius is 3, then the center is at \((5,3)\). Wait, that makes more sense. Let's check the horizontal distance: from the center \(x = 5\) to the rightmost point \(x = 8\), the distance is \(8 - 5=3\), so radius \(r = 3\). And the vertical distance from the center \((5,3)\) to the x - axis (y = 0) is \(3-0 = 3\), which is the radius. So the center \((h,k)=(5,3)\) and radius \(r = 3\).
Now substitute into the standard equation: \((x - 5)^2+(y - 3)^2=3^2\), which is \((x - 5)^2+(y - 3)^2 = 9\). Wait, let's verify. If \(x = 5\), \(y = 3\), then \((5 - 5)^2+(3 - 3)^2=0+0 = 0 eq9\)? No, wait, radius squared is \(r^2\). Wait, no, the radius is 3, so \(r^2=9\). Wait, the center is \((5,3)\), so when \(x = 5\), \(y = 3+3=6\) (top) and \(y = 3 - 3=0\) (bottom), which matches the graph (the bottom is on the x - axis at \(y = 0\)). And horizontally, from \(x = 5\), \(x=5 + 3=8\) (right) and \(x=5-3 = 2\) (left)? Wait, no, the leftmost point is at \(x = 2\)? Wait, the circle is from \(x = 2\) to \(x = 8\) on the x - axis? No, the circle is drawn such that the leftmost point is at \(x = 3\)? Wait, I think I messed up the center. Let's look again. The circle is located with the bottom on the x - axis, centered at \((5,3)\) with radius 3. So the equation is \((x - 5)^2+(y - 3)^2=9\). Wait, let's check the points. For \(x = 5\), \(y = 3\): \((5 - 5)^2+(3 - 3)^2=0 eq9\). Oh no, I made a mistake. The radius squared is \(r^2\), so if the radius is 3, then \(r^2 = 9\), but the center is \((5,3)\), so the equation is \((x - 5)^2+(y - 3)^2=9\). Wait, when \(x = 5\), \(y = 3+3=6\) (top), \(y = 3 - 3=0\) (bottom). When \(x = 5+3=8\), \(y = 3\) (rightmost), \(x = 5 - 3=2\), \(y = 3\) (leftmost). Let's check the graph: the circle is from \(x = 2\) to \(x = 8\) (horizontal) and \(y = 0\) to \(y = 6\) (vertical). So the center is at \((5,3)\), radius 3. So the equation is \((x - 5)^2+(y - 3)^2=9\). Wait, but when I first thought the center was \((5,0)\), that was wrong. The center is at \((5,3)\) because the circle is symmetric above the x - axis, with the bottom at \(y = 0\) and top at \(y = 6\), so the center is at the midpoint of \(y = 0\) and \(y = 6\), which is \(y = 3\), and x - coordinate is 5 (midpoint of \(x = 2\) and \(x = 8\)). So that's correct.

Answer:

\((x - 5)^2+(y - 3)^2 = 9\)