Question 1019885
<pre>
If a circle touches the y-axis, then its radius equals
the x-coordinate of the center. That is, r=h. The center (h,k)
becomes (r,k).  

{{{drawing(400,400,-2,16,-2,16,
circle(2,5,0.15),circle(2,5,0.13),circle(2,5,0.11),circle(2,5,0.09),circle(2,5,0.07),circle(2,5,0.05),circle(2,5,0.03),circle(2,5,0.01),

circle(4,3,0.15),circle(4,3,0.13),circle(4,3,0.11),circle(4,3,0.09),circle(4,3,0.07),circle(4,3,0.05),circle(4,3,0.03),circle(4,3,0.01),

circle(2,3,0.15),circle(2,3,0.13),circle(2,3,0.11),circle(2,3,0.09),circle(2,3,0.07),circle(2,3,0.05),circle(2,3,0.03),circle(2,3,0.01),
green(line(10,11,0,11),line(2,3,0,3), line(2,5,2,3), line(4,3,2,3),
line(10,11,2,3),line(10,11,4,3),line(2,5,10,11)),

locate(10,11,"(r,k)"), 
locate(2,5.8,"(2,5)"),
locate(4.4,3.4,"(4,3)"),
locate(1.7,3,"(r,k)"), 


circle(10,11,0.15),circle(10,11,0.13),circle(10,11,0.11),circle(10,11,0.09),circle(10,11,0.07),circle(10,11,0.05),circle(10,11,0.03),circle(10,11,0.01),






graph(400,400,-2,16,-2,16), circle(2,3,2), circle(10,11,10) )}}} 

So the equation is of the form:

{{{(x-r)^2+(y-k)^2=r^2}}}

The points (2,5) and (4,3) must satisfy this equation, so we have
this system of equations to solve:

{{{system((2-r)^2+(5-k)^2=r^2,(4-r)^2+(3-k)^2=r^2)}}}

Working with the first equation:

{{{(5-k)^2=r^2-(2-r)^2}}}
{{{(5-k)^2=r^2-(4-4r+r^2)}}}
{{{(5-k)^2=r^2-4+4r-r^2}}}
{{{(5-k)^2=-4+4r}}}

Working with the second equation:

{{{(3-k)^2=r^2-(4-r)^2}}}
{{{(3-k)^2=r^2-(16-8r+r^2)}}}
{{{(3-k)^2=r^2-16+8r-r^2}}}
{{{(3-k)^2=-16+8r}}}

We can eliminate r by multiplying the results
of the first equation by -2 and adding to
the results of the second equation:

{{{-2(5-k)^2=8-8r}}}
{{{(3-k)^2=-16+8r}}}

{{{-2(5-k)^2+(3-k)^2=-8}}}
{{{-2(25-10k+k^2)+(9-6k+k^2)=-8}}}
{{{-50+20k-2k^2+9-6k+k^2=-8}}}
{{{-k^2+14k-33 = 0}}}
{{{k^2-14k+33 = 0}}}
{{{(k-3)(k-11)=0}}}
k=3, k=11

Substituting k=3 in
{{{(5-k)^2=-4+4r}}}
{{{(5-3)^2=-4+4r}}}
{{{2^2=-4+4r}}}
{{{4=-4+4r}}}
{{{8=4r}}}
{{{2=r}}}

So the small circle has 
center (2,3) and radius 2, and it has
equation 

{{{(x-2)^2+(y-3)^2=2^2}}}
{{{(x-2)^2+(y-3)^2=4}}}


Substituting k=11 in
{{{(5-k)^2=-4+4r}}}
{{{(5-11)^2=-4+4r}}}
{{{(-5)^2=-4+4r}}}
{{{36=-4+4r}}}
{{{40=4r}}}
{{{10=r}}}

So the large circle has 
center (10,11) and radius 10, and it has
equation 

{{{(x-10)^2+(y-11)^2=10^2}}}
{{{(x-10)^2+(y-11)^2=100}}}

Edwin</pre>