write the equation of the circle that satisfies each set of conditions.

the center of the circle is on the x-axis, its radius is 1, and it passes through the point (square root of 2 over 2, square root of two over 2)

Circle equation generally:

(x-a)^2 + (y-b)^2 = r^2

The center is in x-axis (b = 0), giving:
(x-a)^2 + y^2 = 1
((sqrt(2)/2)-a)^2 + (sqrt(2)/2)^2 = 1
((sqrt(2)/2)-a)^2 + (1/2) = 1
(sqrt(2)/2)-a = ±sqrt(1/2)

(sqrt(2)/2)-a = sqrt(1/2)
a = 0
So: x^2 + y^2 = 1

(sqrt(2)/2)-a = -sqrt(1/2)
a = sqrt(2)
So: (x-sqrt(2))^2 + y^2 = 1

To find the equation of the circle that satisfies the given conditions, we need to understand the equation of a circle.

The equation of a circle with center (h, k) and radius r is given by:

(x - h)^2 + (y - k)^2 = r^2

Let's apply this formula to the given conditions.

Condition 1: The center of the circle is on the x-axis.
When the center is on the x-axis, the value of k is 0, since the y-coordinate of any point on the x-axis is always zero.

Condition 2: The radius of the circle is 1.
The radius (r) of the circle is given as 1.

Condition 3: The circle passes through the point (sqrt(2)/2, sqrt(2)/2).
This means that (sqrt(2)/2, sqrt(2)/2) lies on the circle. We can substitute these coordinates into the equation.

Now let's substitute the given values into the circle equation:

(x - h)^2 + (y - k)^2 = r^2

Since h is the x-coordinate of the center, which lies on the x-axis, h = 0.
Since k is the y-coordinate of the center, which is 0, k = 0.
And r = 1, as given.

Substituting these values into the equation, we get:

(x - 0)^2 + (y - 0)^2 = 1^2
x^2 + y^2 = 1

Therefore, the equation of the circle that satisfies the given conditions is x^2 + y^2 = 1.