Question

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: Identify the center

From the graph, the center of the circle is at \((-1, 5)\) (assuming the grid and axis labels, we analyze the position: looking at the x - axis (vertical axis here? Wait, maybe the axes are labeled with x and y, but the graph's center: let's re - check. Wait, maybe the center is at \((-1, 5)\)? Wait, no, maybe the center is at \((-1, 5)\)? Wait, actually, from the graph, let's see the center coordinates. Let's assume the center is \((h,k)\). Let's check the lowest point on the circle: it touches the origin? Wait, no, the circle seems to have center at \((-1, 5)\)? Wait, maybe I made a mistake. Wait, let's look again. Wait, the vertical axis (x - axis?) and horizontal axis (y - axis?). Wait, maybe the center is at \((-1, 5)\) and radius is 5? Wait, no. Wait, the standard form: let's find the center. Let's see, the circle is centered at \((h,k)\). Let's count the grid squares. Let's assume each grid square is 1 unit. The center of the circle: looking at the graph, the center is at \((-1, 5)\)? Wait, no, maybe the center is at \((-1, 5)\) and the radius is 5? Wait, no, let's calculate the radius. The distance from the center to the origin (if the origin is \((0,0)\))? Wait, no, the circle passes through \((0,0)\)? Wait, the bottom of the circle is at \((0,0)\)? Wait, maybe the center is at \((-1, 5)\) and the radius is 5? Wait, no, let's do it properly.

Wait, let's re - examine the graph. Let's assume the center is \((h,k)\). Let's find two points: the topmost point and the bottommost point. The topmost point seems to be at \((-1, 10)\) and the bottommost at \((-1, 0)\). So the distance between \((-1,10)\) and \((-1,0)\) is \(10 - 0=10\), so the diameter is 10, so the radius \(r = 5\). The center is the mid - point of the vertical line segment from \((-1,0)\) to \((-1,10)\), so the center \((h,k)=(-1,5)\).

Step3: Calculate the radius

As we found, the diameter is 10 (from \(y = 0\) to \(y = 10\) at \(x=-1\)), so the radius \(r = 5\).

Step4: Substitute into the standard equation

Substitute \(h=-1\), \(k = 5\), and \(r = 5\) into \((x - h)^2+(y - k)^2=r^2\).

We get \((x-(-1))^2+(y - 5)^2=5^2\), which simplifies to \((x + 1)^2+(y - 5)^2 = 25\).

Answer:

\((x + 1)^2+(y - 5)^2 = 25\)