Remember that the "shifts" involved with circular functions are sort of like those found in parabolas. When you shift a parabola left or right, you have to think "oppositely". A right shift requires you to subtract from the x-component, and a left one requires you to add. Hence, this circle has a negative 5 horizontal shift, and a negative 22 vertical shift.

You can also remember the general formula for a circle with center at  and a radius of .

Comparing this to the given equation, we can determine the radius and center point.

The center point is at  and the circle has a radius of 11.

The question asks us for the sum of these components:

Remember that the "shifts" involved with circular functions are sort of like those found in parabolas. When you shift a parabola left or right, you have to think "oppositely". A right shift requires you to subtract from the x-component, and a left one requires you to add. Hence, this circle has a positive 50 horizontal shift, and a negative 29 vertical shift.

You can also remember the general formula for a circle with center at  and a radius of .

Comparing this to the given equation, we can determine the radius and center point.

The center point is at  and the circle has a radius of 13.

The question asks us for the sum of these components:

The equation for a circle of radius  centered on point  has the equation 

Just as with linear equations, the horizontal and vertical shifts are opposite of their sign, and are inside the parentheses. The equation of a circle simply must be memorized.

So the circle is centered on point , and plug those in and it yields the formula of the circle.

First determine the center of the circle. The "x-3" portion of the circle equation tells us that the x coordinate is equal to 3. The "y-3" portion of the circle equation tells us that the y coordinate is equal to 3 as well. Therefore, the center of the circle is at (3,3).

To find the distance between (3,3) and (0,0), it is necessary to use the Pythagorean Theorem . Where "a" and "b" are equal to 3

(to visualize, you may draw the two points on a graph, and create a triangle. The line connecting the two points is the hypotenuse, aka "c." )

Write the standard form for the equation of a circle.

The value of  is  and the value of  is .  The center of the circle is:

To find the radius, set  and solve for .

Take the square root of both sides. We only consider the positive value since distance cannot be negative.

The answer is:  

Recall that the standard equation of a circle is . Therefore, in this case . Square root both sides to find your radius. can be simplified to , which is your answer.

First, recall what the standard equation of a circle: . Your center is (h,k). Remember to flip the signs to get your center for this equation: .

Recall what the standard equation of a circle is:

.

is the center of the circle.

Remember that you have to change the signs!

Thus, since our equation is,

your answer for the center is: .

Recall that the standard equation of a circle is

.

Therefore, looking at the equation given,

.

Solve for r to get 7.

Step 1: Recall the general formula of a circle with a center that is not at .

, where  is the center.

Step 2: Find the value of h and k here..

. To find this, we set  of the general equation equal to   (which is in the question)

. To find this, we set  equal to .

The vertex is .

Step 3: To find the radius, take the square root of .

In the equation, we will get .

The radius of the circle is