The center of the circle is .

Find the horizontal distance from the center to the edge of the circle.  At the center , at the edge . The difference is .  This is the radius.

Plug these values:  into the standard form for the equation of a circle.

This gives

.

Remember that the standard form for the equation of a circle is given by the following formula:

Where the point (h,k) gives the center of the circle, and r is the radius. We can see from the form in which the equation is expressed in the problem that the only thing different with our form is that the terms on the left side of the equation are divided by 4. With some algebra, we'll multiply both sides by 4 to eliminate the 4's from the left side of equation:

Now we can see that our equation is the same as the formula for a circle in standard form, where (h,k) is (3,-2) and r=4.

Remember that in order for the equation of a circle to be in standard form, it must be written in the following way:

Where the point (h,k) tells us where the center of the circle is, and r is the radius of the circle. From our answer choices, we can see that the following is the only equation in which there are no fractions and there is addition of terms and not subtraction, which means it is in the standard form shown above:

For this circle, the center would be at (2,-3), and the radius would be 3.

The equation for a circle in general is for a circle with center and radius . We know that the center  is and that one of the points is , so we can determine  by plugging these values in:

Now we can generalize the equation as

Recall that the equation of a circle is  for  the center and  the radius.

In this case, we have  as the center.

Note the negatives in the formula and be careful simpilfying.

When we are done, we have: 

which gives us our answer when simplified. 

To find the standard form of this equation, we have to complete the square for both x and y. It's easiest to do this if we group together the y terms first, then the left terms, and subtract the constant from both sides:

original: subtract 1 from both sides; group x and y

To make the y terms into a square, we have to add 1, since half of 2 is 1, and .

To make the x terms into a square, we have to add 9, since half of -6 is -3, and :

Now we just have to re-write the y and x terms as the squares that they are, and simplify the right side:

To determine the center and radius, put this equation into standard form. Standard form is , where is the center and is the radius.

First, group the y terms together and the x terms together, and subtract 37 from both sides: 

We're leaving a space after these terms because we're completing the square. To make the y terms into a square, we need to add 49, since half of -14 is -7, and

To make the x terms into a square, we need to add 4, since half of -4 is -2, and

We'll need to add these to both sides to keep both sides equal:

Now we can re-write the quadratics on the left as squares, and simplify the right side:

Interpreting this equation, we can see that the center would be at and the radius is

To write this in standard form, we will have to complete the square for both x and y. To do this more easily, group the x terms, then the y terms, and then subtract 43 from both sides:

To complete the square for x, we have to add to both sides; to complete the square for y, we have to add to both sides:

Condense the left side and add together the numbers on the right:

Group the x and y terms on one side of the equation and the constant on the other.

Complete the square by taking half of the middle number for each variable and squaring it. Add the number to the other side of the equation.

The equation of a circle centered at   with radius  units in standard form is

For the circle ceentered at  with radius  units has the equation