To determine the standard-form equation, we'll have to complete the square for both x and y. It will be really helpful to re-group our terms to do that:

Adding 9 will complete the square for x, since

Adding 16 will complete the square for y, since

Now we just need to simplify. Re-write the left side as two binomials squared, and add the numbers on the right side:

Recall that the standard from for the equation of a circle is

where (h, k) is the center, and r is the radius. We are given the center (3, -4) and radius 2. Therefore, h = 3, k = -4, and r = 2. Plugging these vaules into the equation gives us

Since the circle is centered at we use the most basic form for the equation for a circle:

.

Given the circle has a radius of marks, which represent units each, the circle has a radius of units.

We then plug in for :

and simplify: .

Recall the equation of a circle in standard form:

, where (h, k) is the center and r is the radius.

In this problem, we are given the center, but no radius. We must use the other piece of information to find the radius. The second point given is a point on the circle. The definition of a radius is the distance between the center and any point on the circle. Therefore, the radius is equal to the distance between (6,1) and (11,13). Using the distance formula,

Therefore, the radius is 13. Plugging all the information into the standard form of a circle gives us

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:

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.

First we must express the equation in standard form we can determine what the radius of our circle will be. The standard form for the equation of a circle is given as follows:

Where the point (h,k) gives the center of the circle and r is the radius of the circle, which can be easily determined by taking the square root of once the equation is in standard form. Our first step is to multiply both sides of the equation by 3 to cancel the division by 3 on the left side:

Now we can see that our equation is in standard form, where h=-5 and k=4, which tells us the coordinates of the center of the circle:

We can also determine the radius of the circle by taking the square root of :

Now that we know the center of the circle is at (-5,4), and that its radius is 3, we can find the points directly above and below the center by adding 3 to its y-coordinate, and then subtracting 3, respectively, giving us:

and

Similarly, to find the points directly to the left and to the right of the center, we subtract 3 from its x-coordinate, and then add 3, respectively, giving us:

and

We must begin by recalling the general formula for the equation of a circle.

Where circle has center of coordinates and radius of .

That means that looking at our equation, we can see that the center is .

If , then taking the square roots gives us a radius of 2.

We then look at our possible choices. Only two are centered at . Of these two, one has a radius of 2 while one has a radius of 4. We want the former.

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

or

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.