Write the equation that satisfies the stated conditions. There 
may be more than one cirle that satifies the equation. Express the 
final equations in form x^2+y^2+Dx+Ey+F=0

Tangent to the x axis, a radius of length 4, and abscissa of center is -3.
I am so totally lost with this one that I don't even know where to start.

First draw a sketch of such a circle:



So you can see that the center would have to be 4 units above the bottom
point of the circle.  So the center of that circle would have to be
(h, k) = (-3, 4).   So we use the standard form of a circle with center
(h, k) and radius r:

     (x - h)² + (y - k)² = r²

and substitute h = -3, k = 4, and r = 4

     (x + 3)² + (y - 4)² = 4²
     (x + 3)² + (y - 4)² = 16

Then multiply that out and collect terms and
rearrange the equation in the form

     x² + y² + Dx + Ey + F = 0

and you'll get

     x² + y² + 6x - 8y + 9 = 0 
     
Now notice that you could have sketched the circle to hang down
below the x-axis instead of sitting on top of it:



So you can see that the center in this case would have to be 4 units 
below the top point of the circle.  So the center of that circle would 
have to be (h, k) = (-3, -4).   So again we use the standard form of a 
circle with center (h, k) and radius r:

     (x - h)² + (y - k)² = r²

and substitute h = -3, k = -4, and r = 4

     (x + 3)² + (y + 4)² = 4²
     (x + 3)² + (y + 4)² = 16

Then multiply that out and collect terms and
rearrange the equation in the form

     x² + y² + Dx + Ey + F = 0

and you'll get

     x² + y² + 6x + 8y + 9 = 0 

Edwin