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:

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: .

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 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

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

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 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

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:

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: