<h2>
1. What are the center and equation of the circle?</h2>
In this exercise we have the general form of the equation of a circle, which is given by:

The ordinary equation of a circle is:

But this can be written as:

From the original equation we know that:

Finally:

<h2>2. Write the equation of the circle in general form.</h2>
We need to write this equation in the form:

Since:

Then we need to find the center and radius in order to get D, E and F then. From the graph, it is easy to know that the center is
and the radius is 2. Therefore:

Finally, the general form of the equation is:

<h2>3. Write the equation of the parabola</h2>
A parabola is the set of all points in a plane that are equidistant from a fixed line called the directrix and a fixed point called the focus that does not lie on the line. Since the directrix is
then this is a horizontal axis. So the standard for of the equation of a parabola that matches this form is:

The vertex is
so our goal is to find
:

We can find the absolute value of
as follows:

Since the directrix is to the left of the vertex, the parabola opens to the left and hence:
.
Finally, the equation is:
