Given, the distance of point  P(x, y ) from A(– 3, 0) and B(3, 0) is 4 units.

There are two possible solutions for this problem:

1) Solution 1:
To find the distance between two points, we can use the distance formula:

distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

In this case, we have two distances:

distance from (-3,0) to (x,y) = sqrt((x - (-3))^2 + (y - 0)^2)
distance from (3,0) to (x,y) = sqrt((x - 3)^2 + (y - 0)^2)

Since the distance from (-3,0) to (x,y) is 4, we can write the equation:

4 = sqrt((x - (-3))^2 + (y - 0)^2)

Simplifying the equation:

16 = (x + 3)^2 + y^2

Similarly, since the distance from (3,0) to (x,y) is 4, we can write the equation:

4 = sqrt((x - 3)^2 + (y - 0)^2)

Simplifying the equation:

16 = (x - 3)^2 + y^2

Now we have a system of equations:

16 = (x + 3)^2 + y^2
16 = (x - 3)^2 + y^2

Expanding the equations:

16 = x^2 + 6x + 9 + y^2
16 = x^2 - 6x + 9 + y^2

Simplifying the equations:

x^2 + 6x + y^2 = 7
x^2 - 6x + y^2 = 7

Subtracting the second equation from the first equation:

0 = 12x

Solving for x:

x = 0

Substituting x = 0 into one of the equations:

0^2 - 6(0) + y^2 = 7

Simplifying the equation:

y^2 = 7

Taking the square root of both sides:

y = ±sqrt(7)

Therefore, the possible values of x and y are:

x = 0, y = sqrt(7) or x = 0, y = -sqrt(7)

So, the correct answer is:

c) x = 0, y = ±sqrt(7)

2) Solution 2:
We can also solve this problem by graphing the two circles with centers (-3,0) and (3,0) and radius 4.

The equation of the first circle is:

(x + 3)^2 + y^2 = 4^2
(x + 3)^2 + y^2 = 16

The equation of the second circle is:

(x - 3)^2 + y^2 = 4^2
(x - 3)^2 + y^2 = 16

By graphing these two circles, we can see that the only point that satisfies both equations is (0, ±sqrt(7)).

Therefore, the correct answer is:

c) x = 0, y = ±sqrt(7)

Option D is correct