Hello, realintegerz!
1) Find an eq. for a circle that has its center on the yaxis, that has a radius of 4,
and that has only one point on it whose xcoordinate is the same as its ycoordinate Code:

* * *
*  *
*  *
*  *

*  *
* (0,k)* *
*  *

*  *
*  *
*  o P
* * *


       +       

The circle has center (0,k) and radius 4.
It equation is: .$\displaystyle x^2 + (yk)^2 \:=\:16$ .[1]
There is one point $\displaystyle P$ where the xcoordinate and ycoordinate are equal.
. . $\displaystyle \text{That is: }\:x = y.$
Substitute into [1]: .$\displaystyle x^2 + (xk)^2 \:=\:16 \quad\Rightarrow\quad 2x^2  2kx + k^216 \:=\:0$
The Quadratic Formula gives us: .$\displaystyle x \;=\;\frac{k \pm\sqrt{32k^2}}{2}$
It would seem that there are two such points.
Since we are told there is only one such point, the discriminant must be zero.
. . That is: .$\displaystyle 32  k^2 \:=\:0$
Hence: .$\displaystyle k^2 \:=\:32\quad\Rightarrow\quad k \:=\:\pm4\sqrt{2}$
There are two possible circles:
. . $\displaystyle \begin{array}{cc}\text{One above the xaxis:} & x^2 + (y4\sqrt{2})^2 \:=\:16 \\\text{One below the xaxis:} & x^2 + (y+4\sqrt{2})^2 \:=\:16\end{array}$