First plot the point (6,3)



I can see how there could be two different solutions, by drawing in
these:



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



Draw in radii to the axes:



We can see that since the circle has to be tangent to both axes, 
its center has to have the same x and y coordinates. and also that
the radius has to be equal to h as well.  So we can see that all
three values h, k, and r, must all be the same. So let them all be 
h, i.e., h = k = r, and we have



So since , we have


 


Now since it contains the point (6,3) we can substitute that in











That simplifies to



Factoring:



 or 
 or 

So we two values of , so

the two circles' equations are

 and 

 and 



b)The circle is circumscribed about the triangle whose vertices are (-1,-3),(-2,4),and (2,1). 

We plot the points:



We draw the triangle:



The equation of a circle with center (,) and radius  is:



We substitute the point (,) = (,) 













Get all squared terms on right:



We substitute the point (,) = (,) 













Get all squared terms on right:



We substitute the point (,) = (,) 













Get all squared terms on right:



So we have the three equations:




 

Since the right sides of all three equations are equal,
then so are the left sides:



Using the first two





Using the first and third:





So we solve these two equations by 
substitution of elimination:




and get (h,k) = (,)

To find r we go back to



















Therefore the equation 



becomes





So we plot the center (,)



Now put the point of the compass on the center 
and draw the circle:



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Maybe I'll do the last one tomorrow.  I'm getting sleepy. Check back to
see if I've done it.

Edwin