Question 1081620
<pre>
The equation of a circle is

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

Since it's tangent to the y-axis, the center (h,k), is such that
its x-coordinate h = ħr, but since the given points are in the
first quadrant, the circle cannot go into the second quadrant and
be tangent to the y-axis.  So h = r.

so we can substitute r for h

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

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

Factor the left side as the difference of squares:

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

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

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

{{{x^2-2rx=-y^2+2ky+k^2}}}

{{{x^2+y^2-2rx-2ky-k^2=0}}}

Substitute (x,y) = (1,3)

{{{1^2+3^2-2r(1)-2k(3)-k^2=0}}}
 
{{{1+9-2r-6k-k^2=0}}}

{{{10-2r-6k-k^2=0}}}

Solve for 2r, that's the variable I solved for.

{No need to solve for r, for that would induce denominators)

{{{10-6k-k^2=2r}}}

Substitute (x,y) = (2,4)

{{{2^2+4^2-2r(2)-2k(4)-k^2=0}}}
 
{{{4+16-4r-8k-k^2=0}}}

{{{20-4r-8k-k^2=0}}}

Solve for 4r

{{{20-8k-k^2=4r}}}

The system of equations to solve is

{{{system(10-6k-k^2=2r,20-8k-k^2=4r)}}}

Muliply both sides of the first equation by -2
to eliminate the r-terms

{{{system(-20+12k+2k^2=-4r,20-8k-k^2=4r)}}}

Adding the two equations term by term:

{{{4k-k^2=0}}}

{{{k(4-k)=0}}}

k = 0;  4-k = 0
          4 = k

Using k = 0

{{{10-6k-k^2=2r}}}
{{{10-6(0)-0^2=2r}}}
{{{10=2r}}}
{{{5=r}}}

{{{(x-5)^2+(y-0)^2=5^2}}}

This is the graph of the circle with center (h,k)=(5,0) and radius r=5.

{{{drawing(400,400,-10,10,-10,10,graph(400,400,-10,10,-10,10),
circle(5,0,5),circle(5,0,.1), locate(5,1,"(5,0)"),
circle(1,3,.1),circle(2,4,.1),locate(1,3,"(1,3)"),locate(2,4,"(2,4)")  
  )}}}  


Using k = 4

{{{10-6k-k^2=2r}}}
{{{10-6(4)-4^2=2r}}}
{{{10-24-16=2r}}}
{{{-30=2r}}}
{{{-15=r}}}

The radius cannot be negative.  So there is only one solution

Edwin</pre>