Hello, wale!
Did you make any sketches?
$\displaystyle 1.\;\;(x3)^2 +(y2)^2\:=\:36$
Find the equations of the two horizontal tangents to the circle. Code:

 (3,8)
   + * o *   
*  : *
*  : *
*  : r=6 *
 :
*  : *
*  o(3,2) *
*  : *
 :
*+:*
*  : *
*  : *
   + * o *   
 (3,4)

Got it?
$\displaystyle 2.\;\;(x2)^2 +(y2)^2 \:=\:32$
Find the coordinates of the endpoints of the diameter whose equation is $\displaystyle y=x$ Code:

 * * *
*  * *
*  o
*  * *
 *
*  * *
*  o (2,2) *
*  * *
    +          
* *  *
o  *
* *  *
 * * *

We want the intersections of: .$\displaystyle \begin{Bmatrix}[1] & (x2)^2 + (y2)^2 \:=\:32 \\ [2] & x\:=\:y \end{Bmatrix}$
Substitute [2] into [1]: .$\displaystyle (x2)^2 + (x2)^2 \:=\:32 \quad\Rightarrow\quad 2(x2)^2 \:=\:32 $
. . . . . . . $\displaystyle (x2)^2 \:=\:16 \quad\Rightarrow\quad x2 \:=\:\pm4 \quad\Rightarrow\quad x \:=\:2 \pm 4$
Hence: .$\displaystyle x \;=\;6,\:2 \quad\Rightarrow\quad y \;=\;6,\:2$
Therefore, the endpoints of the diameter are: .$\displaystyle (6,6)\,\text{ and }\,(\text{}2,\text{}2)$