Circle O is defined by the equation x2 + (y - 2)2 = 25. Plot the center of Circle O and type in the coordinates of one point with integral values that lies on Circle O.

go to wolframalpha . com to do the plotting. Just type in

implicit plot x^2 + (y-2)^2 = 25

surely you know that 3^2 + 4^2 = 25. So, just find values for x and y such that x^2 = 9 and (y-2)^2 = 16.

(0,2)

To find the center of the Circle O, we can observe that the equation of a circle is in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.

Comparing this with the given equation, x^2 + (y - 2)^2 = 25, we can determine that the center of Circle O is at the point (0, 2).

To find a point on Circle O with integral values, we can substitute some integral values for either x or y into the equation and solve for the other variable.

Let's substitute x = 3 into the equation:
(3)^2 + (y - 2)^2 = 25
9 + (y - 2)^2 = 25
(y - 2)^2 = 16.

Taking the square root of both sides, we get:
y - 2 = ±4.

Solving for y, we have two solutions:
1. y - 2 = 4, so y = 6.
2. y - 2 = -4, so y = -2.

Therefore, two points on Circle O with integral values are (3, 6) and (3, -2).

To plot the center of Circle O, we look at the equation for the circle:

x^2 + (y - 2)^2 = 25.

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

Comparing this with our given equation, we can see that the center of Circle O is (0, 2) since the x-coordinate is 0 and the y-coordinate is 2.

Now, to find a point on the circle with integral values, we can substitute integer values into the equation and solve for the remaining variable.

Let's substitute x = 3 and solve for y:

(3)^2 + (y - 2)^2 = 25
9 + (y - 2)^2 = 25
(y - 2)^2 = 25 - 9
(y - 2)^2 = 16

Taking the square root of both sides, we have:

y - 2 = ±√16
y - 2 = ±4

Solving for y, we get:

y = 2 + 4 = 6 or y = 2 - 4 = -2

So, two points that lie on Circle O and have integral values are (3, 6) and (3, -2).

The center of Circle O is (0, 2), and one point on Circle O with integral values is (3, 6).