The parabola is represented in the form .  If there is a variable in the denominator or as an exponent, it is not a parabola.

The only equation that has an order of two is:   

The equation of a circle with center  and radius  is

The center is at the origin, or , so . To find , use the distance formula as follows:

Note that we do not actually need to find . 

We can now write the equation of the circle:

First, it is necessary to determine the radius of the circle. This is the distance between  and , so we apply the distance formula:

The circumference of the circle is 

Now we need to find the degree measure of the arc. We can do this best by examining this diagram:

The degree measure of  is also the measure  of the central angle formed by the green radii. This is found using the relationship

Using a calculator, we find that . We can adjust for the location of :

which is the degree measure of the arc.

Now we can calculate the length of the arc:

The figure in question is below.

The center of the circle can be seen to be the origin, so, if the radius is , the equation will be

.

The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint  of the segment with endpoints  and  can be found by using the midpoint equations, setting :

The circle passed though this midpoint . The segment from this point to the origin  is a radius, and its length is equal to . Using the following form of the distance formula, since we only need the square of the radius:

,

set :

Substituting in the circle equation for , we get the correct response,

Start by putting the equation in the standard form of the equation of a circle by completing the square. Recall the standard form of the equation of a circle:

, where the center of the circle is at  and the radius is .

From the equation, we know that .

Since the radius is , double its length to find the length of the diameter. The length of the diameter is .

The circumscribed circle of a triangle is the circle which passes through all three vertices of the triangle.

In general form, the equation of a circle is

.

Since the circle passes through the origin, substitute ; the equation becomes

Therefore, we know the equation of any circle passing through the origin takes the form 

for some .

Since the circle passes through , substitute ; the equation becomes

Solving for :

Now we know that the equation takes the form

for some .

Since the circle passes through , substitute ; the equation becomes

Solving for :

The general form of the equation of the circle is therefore

The relation 

is a circle with center  and radius  .

In other words, it is a circle with center at the origin, translated right  units and up  units.

or

is a circle translated right 4 units and up zero units. The upshot is that the circle moves along the -axis only, and therefore is symmetric with respect to the -axis, but not the -axis. Also, as a consequence, it is not symmetric with respect to the origin.

Below is the graph of :

The given graph is the graph of  shifted 6 units left (that is,  unit right) and 3 units up. 

The function graphed is therefore

 where . That is,

Below is the graph of :

If the graph of  is translated by shifting each point  to the point , the graph of

is formed. If the graph is then shifted upward by three units, the new graph is

Since the starting graph was , the final graph is

, or,

Below is the graph of :

If the graph of  is translated by shifting each point  to the point , the graph of

is formed. If the graph is then shifted right by four units, the new graph is

Since the starting graph was , the final graph is

, or