Hello, Anemori!
a. Produce a conic system of inequalities representing the "doughnutshape"
created by an ellipse inside a circle. Code:

..* * *..
*::::::::::*
*::::::::::::::*
*::::::::::::::::*
:::::::::**::::::::
*::::* b *::::* c
  *::*   +   *::*   
*::::*  a *::::*
:::::::***::::::::
*::::::::::::::::*
*::::::::::::::*
*::::::::::*
* * *

The equation of the ellipse is: .$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$
The equation of the circle is: .$\displaystyle x^2 + y^2 \:=\:c^2,\,\text{ where }c \,\geq\,a,b $
System: .$\displaystyle \begin{Bmatrix} \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & \geq & 1 \\ \\[3mm]x^2 + y^2 & \leq & c^2 \end{Bmatrix}$
b. Produce a conic system that has exactly two points of intersection
and consists of a circle and ellipse with the same area.
The equation of an ellipse is: .$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$
. . Its area is: .$\displaystyle \pi ab$
The equation of a circle is: .$\displaystyle x^2 + y^2 \:=\:r^2$
. . Its area is: .$\displaystyle \pi r^2$
The areas are equal: .$\displaystyle \pi ab \:=\:\pi r^2 \quad\Rightarrow\quad ab \:=\:r^2$
Let: .$\displaystyle a = 4,\;b = 1,\;r = 2$
The ellipse is: .$\displaystyle \frac{x^2}{16} + \frac{y^2}{1} \;=\;1$ Code:
1
o o o
o  o
o  o
 o      +      o  
o  o 4
o  o
o o o

The circle is: .$\displaystyle x^2 + y^2 \:=\:4$ Code:
2
◊ ◊ ◊
◊  ◊
◊  ◊

  ◊    +    ◊  
 2
◊  ◊
◊  ◊
◊ ◊ ◊

Move the circle 2 units to the right.
Code:
 :
 ◊ ◊ ◊
o o ♥ : ◊
o ◊ o ◊
o  : o
  o      ◊    +  o  ◊  
o  : o
o ◊ o ◊
o o ♥ :
 ◊ ◊ ◊
System: .$\displaystyle \begin{Bmatrix}\dfrac{x^2}{16} + \dfrac{y^2}{1} \:=\:1 \\ \\[3mm] (x2)^2 + y^2 \:=\:4 \end{Bmatrix}$