SOLUTION: What is the maximum value of k, if the solutions for the two equations below are the same, and both x and y are integers? (x+5)² + (y-1)² = 25 -2x + 3y = k

Algebra ->  Expressions-with-variables -> SOLUTION: What is the maximum value of k, if the solutions for the two equations below are the same, and both x and y are integers? (x+5)² + (y-1)² = 25 -2x + 3y = k      Log On


   



Question 1171717: What is the maximum value of k, if the solutions for the two equations below are the same, and both x and y are integers?
(x+5)² + (y-1)² = 25
-2x + 3y = k

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
system%28%28x%2B5%29%5E2+%2B+%28y-1%29%5E2+=+25%2C-2x+%2B+3y+=+k%29

Put the second in slope-intercept form:

-2x+%2B+3y+=+k
3y+=+2x%2Bk
y+=+expr%282%2F3%29x%2Bk%2F3

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



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:



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 (x1,y1) to the line Ax + By + C = 0

is given by the formula: 

d=abs%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29

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

abs%28-2%28-5%29%2B3%281%29-k%29%2Fsqrt%28%28-2%29%5E2%2B%283%29%5E2%29=5

abs%2810%2B3-k%29%2Fsqrt%284%2B9%29=5

abs%2813-k%29%2Fsqrt%2813%29=5

abs%2813-k%29=5sqrt%2813%29

13-k=5sqrt%2813%29 or -13%2Bk=5sqrt%2813%29
-k=-13%2B5sqrt%2813%29 or k=13%2B5sqrt%2813%29
k=13-5sqrt%2813%29 

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

So the maximum value of k is

k=13%2B5sqrt%2813%29 which is approximately 31.02775638

The largest potential integer k is 31.

And it does turn out that when k=31

system%28%28x%2B5%29%5E2+%2B+%28y-1%29%5E2+=+25%2C-2x+%2B+3y+=+31%29

We get two solutions (x,y) = (-8,5) and (x,y) = %28matrix%281%2C3%2C-98%2F13%2C%22%2C%22%2C69%2F13%29%29

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

Edwin