SOLUTION: A circle with a radius of 4 and a midpoint at (2,2)

It is unusual to call the center (or "centre" if you are from the UK)
of a circle by the term "midpoint".  We usually use the word "midpoint"
for the midpoint of a line segment, or for the midpoint between two 
given points.  But I will take it that by "midpoint", you mean "center"
or "centre".

So I assume you want the equation of this green circle, with the center 
at (2,2), the point so marked below:

 

Then you must learn the standard form for the equation of a
circle which has center at (h,k) and radius r:



If you know that formula by heart, then all you have to do is
substitute h=2, k=2 since (h,k) = (2,2) and r=4, and instantly
you have



That's it except for the tiny task of squaring the 4



except that it is better not to acrtually square the
radius.  Why? The standard form was created because
they wanted to get equations in forms so that you can
tell how the graph is going to look just by looking at 
the equation.

You don't need to multiply that out at all. Just leave it
in standard form.  The beauty of the standard form is that
you can look at the equation in that form and tell about
the graph easier. You can just look at the equation and
pick out the radius and the center without doing any
algebra.   



The two red 2's tell you the center is (2,2) and the green 4 tells you
that the radius is 4, so you can draw it in your head.

On top of that, it IS the equation of that circle because the
points (-2,2), (2,-2), (2,6), (6,2) all lie on this circle
as you can see:



Plug each into 



Plugging in (-2,2)






Plugging in (2,-2)








Plugging in (2,6)






Plugging in (6,2)






So you see that all those four sample points
satisfy the equation in standard form. That
convinces us that it must be the right
equation for that circle.

Edwin