Question 990255
Unknown center of the first circle, (h,k).
Using Distance Formula
{{{(h-8)^2+(k-7)^2=(h-0)^2+(k-3)^2}}}
{{{(h-8)^2+(k-7)^2=h^2+(k-3)^2}}}
{{{h^2-16h+64+k^2-14k+49=h^2+k^2-6k+9}}}
{{{-16h+64-14k+49=-6k+9}}}
{{{-16h-14k+6k+64+49-9=0}}}
{{{-16h-8k+64+40=0}}}
{{{16h+8k=104}}}
{{{2h+k=13}}}


This circle touching, just touching, the y-axis at (0,3) means that the center is some point on the line y=3.  This means, you can solve for h, because you know  {{{k=-3}}}.  This is based on knowing how the standard circle equation works.


{{{2h=13-k}}}
{{{2h=13-(-3)}}}
{{{2h=16}}}
{{{h=8}}}


Center of this first circle is therefore,  (8,-3).  The point (0,3) on the circle may be the convenient point to again use the Distance Formula, to find the radius of this circle.


{{{r^2=(8-0)^2+(-3-0)^2}}}
{{{r^2=8^2}}}
{{{r=8}}}


This first circle equation is then,  {{{highlight((x-8)^2+(y+3)=8)}}}.
I have not finished to do the final question, but maybe you can.