Question 1171717
<pre>
{{{system((x+5)^2 + (y-1)^2 = 25,-2x + 3y = k)}}}

Put the second in slope-intercept form:

{{{-2x + 3y = k}}}
{{{3y = 2x+k}}}
{{{y = expr(2/3)x+k/3}}}

The first equation is a circle and the second represents a set of parallel lines with slope 2/3 and y-intercept k/3.

{{{drawing(400,400,-12,12,-12,12,  graph(400,400,-12,12,-12,12),circle(-5,1,.1), 
circle(-5,1,5),line(-12,-8,12,8),       line(-14,4,13,22),
line(-12,2,12,18), line(-13,-2,14,16),line(-11,-4,13,12),line(-13,-12,14,6)

 )}}}

We want to find the maximum value of k.  k will be a maximum when k/3 is
a maximum. k/3 is the y-intercept and it will be a maximum when the y-intercept
is the largest.  That will be when the line is the upper tangent to the circle.
Like this:

{{{drawing(400,400,-12,12,-12,12,  graph(400,400,-12,12,-12,12),circle(-5,1,.1),  
circle(-5,1,5),line(-12,2.342585459,12,18.34258546)
 )}}}

We know the distance from the center (-5,1) to the point of tangency is the
radius, which is 5. So we use the formula for the perpendicular distance
from a point to a line, which is

The perpendicular distance from the point (x<sub>1</sub>,y<sub>1</sub>) to the line Ax + By + C = 0

is given by the formula: 

{{{d=abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

So the perpendicular distance from the center (-5,1) to the line
-2x + 3y - k = 0 is equal to the radius 5, so

{{{abs(-2(-5)+3(1)-k)/sqrt((-2)^2+(3)^2)=5}}}

{{{abs(10+3-k)/sqrt(4+9)=5}}}

{{{abs(13-k)/sqrt(13)=5}}}

{{{abs(13-k)=5sqrt(13)}}}

{{{13-k=5sqrt(13)}}} or {{{-13+k=5sqrt(13)}}}
{{{-k=-13+5sqrt(13)}}} or {{{k=13+5sqrt(13)}}}
{{{k=13-5sqrt(13)}}} 

The first is the lower tangent line, so we
want the upper tangent line,

So the maximum value of k is

{{{k=13+5sqrt(13)}}} which is approximately 31.02775638

The largest potential integer k is 31.

And it does turn out that when k=31

{{{system((x+5)^2 + (y-1)^2 = 25,-2x + 3y = 31)}}}

We get two solutions (x,y) = (-8,5) and (x,y) = {{{(matrix(1,3,-98/13,",",69/13))}}}

So x=-8, y=5, is an integer solution that allows k to be the maximum integer
value of 31. 

Edwin</pre>