Question

write the standard equation for each of the circles in parts (a) through (a). the coordinates of the center and the radius for each circle are integers. (a) the equation of the circle in standard form is ( x^2 + y^2 = 9 ). (type an equation. simplify your answer.) (b) the equation of the circle in standard form is ( x^2 + y^2 = ). (type an equation. simplify your answer.)

Explanation:

Part (a) (already given as \(x^2 + y^2 = 9\), but let's verify)

Step1: Recall circle standard form

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 center and radius

From the graph, the center is at \((0, 0)\) (since it's the origin, as the circle is symmetric about both axes and the center is at the intersection of the axes). The radius can be found by the distance from the center \((0,0)\) to a point on the circle, like \((3, 0)\). Using the distance formula \(r=\sqrt{(3 - 0)^2+(0 - 0)^2}=\sqrt{9}=3\).

Step3: Substitute into standard form

Substitute \(h = 0\), \(k = 0\), and \(r = 3\) into \((x - h)^2 + (y - k)^2 = r^2\). We get \((x - 0)^2+(y - 0)^2=3^2\), which simplifies to \(x^2 + y^2 = 9\).

Part (b)

Step1: Recall circle standard form

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 center and radius

From the graph, the center is at \((3, 0)\) (marked as "Center" at \((3,0)\)). A point on the circle is \((1, 0)\). To find the radius, we calculate the distance between the center \((3, 0)\) and the point \((1, 0)\) on the circle. Using the distance formula \(r=\sqrt{(3 - 1)^2+(0 - 0)^2}=\sqrt{4}=2\).

Step3: Substitute into standard form

Substitute \(h = 3\), \(k = 0\), and \(r = 2\) into \((x - h)^2 + (y - k)^2 = r^2\). We get \((x - 3)^2+(y - 0)^2=2^2\), which simplifies to \((x - 3)^2 + y^2 = 4\). If we expand \((x - 3)^2=x^2-6x + 9\), then \(x^2-6x + 9+y^2=4\), or \(x^2+y^2=4 + 6x-9=6x - 5\)? Wait, no, the problem seems to have a typo in the initial box for part (b), but following the standard form, since the center is \((3,0)\) and radius \(2\), the standard form is \((x - 3)^2 + y^2 = 4\). But if we want to write it in the form \(x^2 + y^2=\dots\), we expand \((x - 3)^2 + y^2 = 4\) to \(x^2-6x + 9+y^2=4\), so \(x^2 + y^2=6x - 5\). But maybe the problem expects the standard form \((x - h)^2+(y - k)^2=r^2\). Wait, looking at the graph again, maybe the center is \((3,0)\) and the point on the circle is \((1,0)\), so radius is \(2\). So the standard equation is \((x - 3)^2 + y^2 = 4\). But if we consider the form \(x^2 + y^2=\dots\), expanding gives \(x^2 + y^2=6x - 5\). But perhaps there's a mistake in my reading. Wait, maybe the center is \((3,0)\) and the point is \((1,0)\), so radius \(r = 2\). So the standard equation is \((x - 3)^2 + y^2 = 4\).

Answer:

(for part b):
The equation of the circle in standard form is \(\boldsymbol{(x - 3)^2 + y^2 = 4}\) (or if we expand to \(x^2 + y^2\) form: \(x^2 + y^2=6x - 5\), but the standard form is \((x - 3)^2 + y^2 = 4\)).