The parabola is represented in the form $\displaystyle y=ax^2+bx+c$.  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:   $\displaystyle y=\frac{1}{2}x^2$

The equation of a circle with center $\displaystyle (h, k)$ and radius $\displaystyle r$ is

$\displaystyle (x-h)^{2}+ \left ( y - k\right )^{2} = r^{2}$

The center is at the origin, or $\displaystyle (0,0)$, so $\displaystyle h = k = 0$. To find $\displaystyle r^{2}$, use the distance formula as follows:

$\displaystyle r^{2} = \left ( 6-0\right )^{2} +\left ( 8 -0 \right )^{2} = 6^{2}+8^{2} = 36 + 64 = 100$

Note that we do not actually need to find $\displaystyle r$. 

We can now write the equation of the circle:

$\displaystyle (x-0)^{2}+ \left ( y - 0\right )^{2} = 100$

$\displaystyle x^{2}+y^{2} = 100$

First, it is necessary to determine the radius of the circle. This is the distance between $\displaystyle (0,0)$ and $\displaystyle (-9, -12)$, so we apply the distance formula:

$\displaystyle r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}$

$\displaystyle r = \sqrt{\left ( -9-0\right )^{2} +\left ( -12 -0 \right )^{2} }$

$\displaystyle r=\sqrt{ (-9)^{2}+(-12)^{2}}$

$\displaystyle r= \sqrt{81+144 }= \sqrt{225} = 15$

The circumference of the circle is 

$\displaystyle C = 2 \pi r = 2 \pi \cdot 15 = 30 \pi$

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

The degree measure of $\displaystyle \widehat{AB}$ is also the measure $\displaystyle \theta$ of the central angle formed by the green radii. This is found using the relationship

$\displaystyle \tan \theta = \frac{y}{x}$

$\displaystyle \tan \theta = \frac{-12}{-9} \approx 1.333$

Using a calculator, we find that $\displaystyle \tan ^{-1} 1.3333 \approx 53.12^{\circ }$. We can adjust for the location of $\displaystyle B$:

$\displaystyle \theta \approx \left (180 - 53.12 \right )^{\circ } \approx 126.87^{\circ }$

which is the degree measure of the arc.

Now we can calculate the length of the arc:

$\displaystyle L \approx \frac{126.87}{360} \cdot C$

$\displaystyle L \approx \frac{126.87}{360} \cdot 30 \pi \approx 33.2$

The figure in question is below.

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

$\displaystyle x^{2}+ y^{2} = r^{2}$.

The circle passes through the midpoints of the sides, so we will find one of these midpoints. The midpoint $\displaystyle (x_{m}, y_{m})$ of the segment with endpoints $\displaystyle (7,0)$ and $\displaystyle (0, 7)$ can be found by using the midpoint equations, setting $\displaystyle x_{1} = y_{2}= 7, x_{2} = y_{1}= 0$:

$\displaystyle x_{m} = \frac{x_{1}+x_{2}}{2} = \frac{7+0}{2} = \frac{7}{2}$

$\displaystyle y_{m} = \frac{y_{1}+y_{2}}{2} = \frac{0+7 }{2} = \frac{7}{2}$

The circle passed though this midpoint $\displaystyle \left ( \frac{7}{2}, \frac{7}{2} \right )$. The segment from this point to the origin $\displaystyle (0,0)$ is a radius, and its length is equal to $\displaystyle r$. Using the following form of the distance formula, since we only need the square of the radius:

$\displaystyle r ^{2} = (x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}$,

set $\displaystyle x_{1} = y_{1} = 0, x_{2}= y_{2}= \frac{7}{2}$:

$\displaystyle r ^{2} =\left (\frac{7}{2}-0\right ) ^{2}+\left (\frac{7}{2}-0\right ) ^{2}$

$\displaystyle =\left (\frac{7}{2} \right ) ^{2}+\left (\frac{7}{2}\right ) ^{2}$

$\displaystyle = \frac{49}{4} + \frac{49}{4}$

$\displaystyle = \frac{49}{2}$

Substituting in the circle equation for $\displaystyle r ^{2}$, we get the correct response,

$\displaystyle x^{2}+ y^{2}= \frac{49}{2}$

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:

$\displaystyle (x-a)^2+(y-b)^2=r^2$, where the center of the circle is at $\displaystyle (a,b)$ and the radius is $\displaystyle r$.

$\displaystyle x^2-10x+25+y^2+2y+1+10=25+1$

$\displaystyle (x-5)^2+(y+1)^2+10=26$

$\displaystyle (x-5)^2+(y+1)^2=16$

From the equation, we know that $\displaystyle r^2=16$.

$\displaystyle r=\sqrt{16}=4$

Since the radius is $\displaystyle 4$, double its length to find the length of the diameter. The length of the diameter is $\displaystyle 8$.

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

$\displaystyle x^{2}+ y^{2}+ Ax + By + C = 0$.

Since the circle passes through the origin, substitute $\displaystyle x = 0, y = 0$; the equation becomes

$\displaystyle 0^{2}+0^{2}+ A \cdot 0 + B \cdot 0 + C = 0$

$\displaystyle 0+0+0+0 + C = 0$

$\displaystyle C= 0$

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

$\displaystyle x^{2}+ y^{2}+ Ax + By = 0$

for some $\displaystyle A , B$.

Since the circle passes through $\displaystyle (7,0)$, substitute $\displaystyle x = 7, y = 0$; the equation becomes

$\displaystyle 7^{2}+ 0^{2}+ A \cdot 7 + B \cdot 0 = 0$

$\displaystyle 49+7A= 0$

Solving for $\displaystyle A$:

$\displaystyle 7A = -49$

$\displaystyle A = -7$

Now we know that the equation takes the form

$\displaystyle x^{2}+ y^{2}-7x + By = 0$ 

for some $\displaystyle B$.

Since the circle passes through $\displaystyle (4, 4)$, substitute $\displaystyle x = 7, y = 0$; the equation becomes

$\displaystyle 4^{2}+ 4^{2}- 7 \cdot 4 + B \cdot 4 = 0$

$\displaystyle 16+16- 28 + 4B = 0$

$\displaystyle 4 + 4B = 0$

Solving for $\displaystyle B$:

$\displaystyle 4B = -4$

$\displaystyle B = -1$

The general form of the equation of the circle is therefore

$\displaystyle x^{2}+ y^{2}-7x -y = 0$

The relation 

$\displaystyle (x-h)^{2} + (y - k)^{2} = r^{2}$

is a circle with center $\displaystyle (h,k)$ and radius $\displaystyle r$ .

In other words, it is a circle with center at the origin, translated right $\displaystyle h$ units and up $\displaystyle k$ units.

$\displaystyle \left (x- 4 \right )^{2} + y ^{2}= 18$

or

$\displaystyle \left (x- 4 \right )^{2} +\left ( y - 0 \right ) ^{2}= 18$

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

Below is the graph of $\displaystyle f(x) = |x|$:

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

The function graphed is therefore

$\displaystyle g (x) = f(x-h) + k$ where $\displaystyle h = 6, k= 3$. That is,

$\displaystyle g (x) = f(x-6) +3$

$\displaystyle g(x)= | x-6 | + 3$

Below is the graph of $\displaystyle f(x) = |x|$:

If the graph of $\displaystyle f$ is translated by shifting each point $\displaystyle (x,y)$ to the point $\displaystyle (2, 2y)$, the graph of

$\displaystyle g(x) = 2f(x)$ 

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

$\displaystyle g(x) = 2f(x) + 3$

Since the starting graph was $\displaystyle f(x) = |x|$, the final graph is

$\displaystyle g(x) = 2 |x| + 3$, or,

$\displaystyle g(x) = |2x| + 3$

Below is the graph of $\displaystyle f(x) = |x|$:

If the graph of $\displaystyle f$ is translated by shifting each point $\displaystyle (x,y)$ to the point $\displaystyle (x, 3y)$, the graph of

$\displaystyle g(x) = 3f(x)$ 

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

$\displaystyle g(x) = 3f(x- 4)$

Since the starting graph was $\displaystyle f(x) = |x|$, the final graph is

$\displaystyle g(x) = 3 |x-4|$, or $\displaystyle g(x) = |3x-12|$