Question

write the standard equation for each of the circles in parts (a) through (c). 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 = 4 ). (type an equation. simplify your answer.)
(b) the equation of the circle in standard form is \\(\square\\). (type an equation. simplify your answer.)

Explanation:

Part (b) (Assuming the circle with center (6,0) and passing through (1,6))

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. Here, the center \((h,k)=(6,0)\).

Step2: Calculate the radius

The radius \(r\) is the distance between the center \((6,0)\) and the point \((1,6)\) on the circle. Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we have:
\(r=\sqrt{(1 - 6)^2+(6 - 0)^2}=\sqrt{(-5)^2+6^2}=\sqrt{25 + 36}=\sqrt{61}\)

Step3: Write the equation

Substitute \(h = 6\), \(k = 0\), and \(r^2 = 61\) into the standard form:
\((x - 6)^2+(y - 0)^2 = 61\), which simplifies to \((x - 6)^2 + y^2 = 61\)

Part (b) (Assuming the circle with center at the midpoint? Wait, the other circle has points (-2,0), (2,0), (0,2), (0,-2). Let's check that circle:

Step1: Find the center

The center is the midpoint of the horizontal and vertical diameters. For horizontal points \((-2,0)\) and \((2,0)\), midpoint is \((0,0)\)? Wait no, wait the points are \((-2,0)\), \((2,0)\), \((0,2)\), \((0,-2)\). Wait, the center should be the intersection of the perpendicular bisectors. The horizontal diameter is from \((-2,0)\) to \((2,0)\), midpoint is \((0,0)\), and vertical diameter from \((0,-2)\) to \((0,2)\), midpoint is \((0,0)\). So center \((h,k)=(0,0)\)? Wait no, wait the distance from center to \((2,0)\) is radius. Wait, no, the points are \((-2,0)\), \((2,0)\), \((0,2)\), \((0,-2)\). So the center is \((0,0)\)? Wait, no, wait the horizontal distance from center to \((2,0)\) is 2, vertical distance to \((0,2)\) is 2. Wait, but the standard form: if center is \((0,0)\), radius 2? But no, wait the distance from \((0,0)\) to \((2,0)\) is 2, to \((0,2)\) is 2. Wait, but the equation would be \(x^2 + y^2 = 4\)? No, wait no, wait the points: \((-2,0)\) is on the circle, so \((-2 - h)^2+(0 - k)^2 = r^2\), \((2 - h)^2+(0 - k)^2 = r^2\), \((0 - h)^2+(2 - k)^2 = r^2\), \((0 - h)^2+(-2 - k)^2 = r^2\). Solving these, we get \(h = 0\), \(k = 0\), and \(r^2 = 4\)? Wait no, \((2 - 0)^2+(0 - 0)^2 = 4\), so \(r^2 = 4\), so equation \(x^2 + y^2 = 4\)? But that's part (a). Wait, maybe the other circle: the one with center (6,0) and passing through (1,6). Let's confirm:

Wait, the image shows two circles: one with center (6,0) and a point (1,6) on it, and another with points (-2,0), (2,0), (0,2), (0,-2) (center at (0,0)? Wait no, the center of the second circle: the horizontal line through (-2,0) and (2,0) is the x-axis, vertical line through (0,2) and (0,-2) is the y-axis, so center is (0,0), radius is 2 (distance from (0,0) to (2,0) is 2), so equation \(x^2 + y^2 = 4\) (which is part (a)). Then the first circle: center (6,0), point (1,6) on it. So radius squared is \((1 - 6)^2 + (6 - 0)^2 = 25 + 36 = 61\), so equation \((x - 6)^2 + y^2 = 61\).

Assuming part (b) is the circle with center (6,0) and passing through (1,6):

Answer:

\((x - 6)^2 + y^2 = 61\)

(If part (b) is the other circle, but that's part (a)'s equation. So likely the first circle. If there's a different center, please clarify, but based on the image, the circle with center (6,0) and point (1,6) gives the equation above.)