Question 1086693
<pre>
There are obviously two solutions.

{{{drawing(800,800,-6,32,-6,32,
green(line(-3,2,-2,5),line(9,-2,14,13),line(-2,5,1,4),line(-2,5,-1,8),
line(14,13,1,4),line(14,13,-1,8)),
locate(-4,5.5,"(h,k)"),locate(14,13,"(h,k)"),
locate(1,4,"A(1,4)"), locate(-3.5,8.8,"B(-1,8)"),
graph(800,800,-6,32,-6,32), line(-12,5,33,-10),

circle(-2,5,sqrt(10)), circle(14,13,5sqrt(10)) )}}}

We set the three radii equal. We use the distance formula
for two of them and the formula for the perpendicular
distance from a point to a line for the radius to the
point of tangency.

{{{sqrt((h+1)^2+(k-8)^2)=sqrt((h-1)^2+(k-4)^2)=abs(h+3k-3)/sqrt(1^2+3^2)}}}

{{{sqrt((h+1)^2+(k-8)^2)=sqrt((h-1)^2+(k-4)^2)=abs(h+3k-3)/sqrt(10)}}}

Square all three 

{{{(h+1)^2+(k-8)^2=(h-1)^2+(k-4)^2 = (h+3k-3)^2/10}}}

Taking the first two

{{{(h+1)^2+(k-8)^2=(h-1)^2+(k-4)^2}}}

That simplifies down to 

{{{h=2k-12}}}

Substitute 2k-12 for h in the second and third

{{{(h-1)^2+(k-4)^2 = (h+3k-3)^2/10}}}

{{{(2k-12-1)^2+(k-4)^2 = (2k-12+3k-3)^2/10}}}

That simplifies to 

{{{(k-5)(x-13)=0}}}

So k = 5 or k = 13

Substituting in

{{{h=2k-12}}}

h = -2 or h = 14

the centers of the two circles are (-2,5) and (14,13) 

To find the radius, substitute in any one of the 
three expressions for the radius, say, the third one

{{{r=abs(h+3k-3)/sqrt(10)}}}
{{{abs(-2+3(5)-3)/sqrt(10)}}}
{{{10/sqrt(10)}}}
{{{10sqrt(10)/10}}}
{{{sqrt(10)}}}

The standard equation of a circle is:

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

So the smaller circle has equation

{{{(x+2)^2+(y-5)^2=10}}}

For the larger circle:

{{{r=abs(h+3k-3)/sqrt(10)}}}
{{{abs(14+3(13)-3)/sqrt(10)}}}
{{{50/sqrt(10)}}}
{{{50sqrt(10)/10}}}
{{{5sqrt(10)}}}

So the larger circle has equation

{{{(x-14)^2+(y-13)^2=250}}}

Edwin</pre>