Question 1099107
<pre>
Here's a better way.

{{{drawing(400,400,-9,9,-4,14, graph(400,400,-9,9,-4,14),
circle(-2/7,41/7,10sqrt(13)/7), line(-14,-24,12,15),locate(-2.5,6.3,"(h,k)"),
locate(4,3,"(4,3)"), circle(-2/7,41/7,.2), green(line(4,3,-2/7,41/7),

line(-2,1,-2/7,41/7)),  locate(-4.3,1,"(-2,1)"),circle(4,3,.2), circle(-2,1,.2)


 )}}}

We set the lengths of the two green radii equal:

{{{sqrt((h-4^"")^2+(k-3^"")^2)}}}{{{""=""}}}{{{sqrt((h-(-2)^"")^2+(k-1^"")^2)}}}

That simplifies to:

{{{3h+k}}}{{{""=""}}}{{{5}}}

That's one equation in h and k.

To get the other equation in h and k, we use the fact that a 
tangent to a circle is perpendicular to the radius drawn to the 
point of tangency.

We find the slope of the tangent line:

{{{3x-2y-6}}}{{{""=""}}}{{{0}}}
{{{-2y}}}{{{""=""}}}{{{-3x+6}}}
{{{(-2y)/(-2)}}}{{{""=""}}}{{{expr((-3)/(-2))x+6/(-2)}}}
{{{y}}}{{{""=""}}}{{{expr(3/2)x-3}}}

Comparing that to y = mx+b, the tangent line has slope
{{{3/2}}} so the radius drawn to the point of tangency
has slope {{{-2/3}}}, its "negative reciprocal".

So we use the slope formula and set its slope equal to {{{-2/3}}}

{{{(k-3)/(h-4)}}}{{{""=""}}}{{{-2/3}}}

That simplifies to:

{{{2h+3k}}}{{{""=""}}}{{{17}}}

So we solve this system:

{{{system(3h+k=5,2h+3k=17)}}}

Solve that by substitution or elimination and get

{{{(matrix(1,3,h,",",k))}}}{{{""=""}}}{{{(matrix(1,3,-2/7,",",41/7))}}}

Then we find the length of the radius, using the distance
formula for the length of the radius drawn to the point 
of tangency:

{{{sqrt((h-4^"")^2+(k-3^"")^2)}}}
{{{sqrt((-2/7-4^"")^2+(41/7-3^"")^2)}}}

which simplifies to {{{10sqrt(13)/7}}}

Substituting for the center and radius in

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

{{{(x-(-2/7)^"")^2+(y-(41/7)^"")^2}}{{{""=""}}}{{{(10sqrt(13)/7)^2}}}

which simplifies to:

{{{(x + 2/7)^2 + (y - 41/7)^2}}}{{{""=""}}}{{{1300/49}}}

Edwin</pre>