Question 1191933
<pre>
A line tangent to a circle is perpendicular to the radius drawn to
the point of tangency.  So the radius drawn to the point of tangency
will lie along the line perpendicular to the line 5x + y = 3 that 
passes through the point of tangency (2,-7) 


{{{drawing(400,400,-4,14,-14,4,
circle(7,-6,sqrt(26)),

line(-18,-11,22,-3),
graph(400,400,-4,14,-14,4),line(-5,28,15,-72),line(-5,-12,15,-2)

)}}}

We find the equation of the line perpendicular to the line 5x + y = 3 at the
point (2,-7). The line 

5x+y=3
   y=-5x+3

has slope -5, so a line perpendicular to it has slope which is the negative
reciprocal of -5 which is +1/5.  So we find the equation of the line
through (2,-7) with slope 1/5:

{{{y-y[1]}}}{{{""=""}}}{{{m(x-x[1])}}}
{{{y-(-7)}}}{{{""=""}}}{{{expr(1/5)(x-2)}}}
{{{y+7)}}}{{{""=""}}}{{{expr(1/5)(x-2)}}}
{{{5y+35)}}}{{{""=""}}}{{{x-2}}}
{{{37}}}{{{""=""}}}{{{x-5y}}}
{{{x-5y}}}{{{""=""}}}{{{37}}}

We now know that the center lies on the line x - 5y = 37, and we
are given that the center lies on the line x - 2y = 19.  So those 
lines must intersect at the center of the circle, so we solve them
simultaneously to find their intersection, which is the center of
the circle.

{{{system(x-5y=37,x-2y=19)}}}

Solve those by substitution or elimination and get x=7, y=-6
So they intersect at (7,-6), the center of the circle.

To find the radius, all we need do is find the distance from the
center (7,-6) to the point of tangency (2,-7).  We use the distance
formula:

{{{d}}}{{{""=""}}}{{{sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

{{{d}}}{{{""=""}}}{{{sqrt(((2)^""-(7))^2+((-7)-(-6)^"")^2)}}}

{{{d}}}{{{""=""}}}{{{sqrt((2-7)^2+(-7+6)^2)}}}

{{{d}}}{{{""=""}}}{{{sqrt((-5)^2+(-1)^2)}}}

{{{d}}}{{{""=""}}}{{{sqrt(25+1)}}}

{{{d}}}{{{""=""}}}{{{sqrt(26)}}}

So the radius is {{{sqrt(26)}}} and the center is (7,-6).

The equation of the circle is

{{{(x-7^"")^2+(y-(-6)^"")^2}}}{{{""=""}}}{{{(sqrt(26))^2}}}

{{{(x-7^"")^2+(y+6^"")^2}}}{{{""=""}}}{{{26}}}

Edwin</pre>