SOLUTION: If the solutions for the two equations below are the same, find the maximum value of k, if: <img src="https://i.ibb.co/Y7TxqYQ/1.png" height="25px"> Here are the two equation

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: If the solutions for the two equations below are the same, find the maximum value of k, if: <img src="https://i.ibb.co/Y7TxqYQ/1.png" height="25px"> Here are the two equation      Log On


   

Question 1205414: If the solutions for the two equations below are the same, find the maximum value of k, if:

Here are the two equations:
Found 4 solutions by Edwin McCravy, MathLover1, ikleyn, mccravyedwin:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
system%28%28x%2B2%29%5E2%2B%28y-5%29%5E2=4%2C-4x%2By=k%29
y=4x%2Bk



That's a circle with center (-2,5) and radius 2 intersecting a line with
slope 4 and y-intercept k. So, k, the y-intercept of the line will be the
largest when the line is as far to the left of the circle and still intersects
the circle.



That's when the line is tangent to the circle on the left side. That's also 
when the distance from the line to the center is the radius 2.

We use the point-to-line distance formula:

We get 0 on the right side of the line's equation:

4x-y%2Bk=0

Perpendicular distance from the point (x1,y1)
to the line Ax+By+C=0 is

d%22%22=%22%22abs%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29

2%22%22=%22%22abs%284%28-2%29-5%2Bk%29%2Fsqrt%284%5E2%2B%28-1%29%5E2%29

2%22%22=%22%22abs%28-8-5%2Bk%29%2Fsqrt%2816%2B1%29

2%22%22=%22%22abs%28-13%2Bk%29%2Fsqrt%2817%29

2sqrt%2817%29%22%22=%22%22abs%28-13%2Bk%29

-13%2Bk%22%22=%22%22%22%22+%2B-+2sqrt%2817%29
k%22%22=%22%2213+%2B-+2sqrt%2817%29
We want k to be a maximum so we use the + sign.

k%22%22=%22%2213+%2B+2sqrt%2817%29 approximately 21.24621125.

Oh darn, I just realized that x and y had to be integers. Instead of
starting over it looks like the nearest integer x could be -4. So we have
to move the line a tiny bit right.

So, we plug -4 for x in the circle equation:

%28-4%2B2%29%5E2%2B%28y-5%29%5E2=4
%28-2%29%5E2%2B%28y-5%29%5E2=4
4+%2B+%28y-5%29%5E2=4
%28y-5%29%5E2=0
y=5

So (-4,5) is the nearest point to the point of tangency that
has both coordinates integers.
So the line y=4x+k should go through (-4,5), moving it a tiny
bit right.

y=4x%2Bk
5=4%28-4%29%2Bk
5=-16%2Bk
21=k

So the answer is 21.
 
Edwin

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

%28x%2B2%29%5E2+%2B+%28y-5%29%5E2=4.....eq.1
-4x%2By=k......eq.2
----------------------
start with
%28x%2B2%29%5E2+%2B+%28y-5%29%5E2=4.....eq.1, this is a circle with center at (-2,5) and radius r=2

and find intercepts
the y-intercepts can be found by setting x=0+in the equation and solving for y%7D%5D%7D%3A%0D%0A%0D%0A%7B%7B%7B%280%2B2%29%5E2+%2B+%28y-5%29%5E2=4+
4+%2B+%28y-5%29%5E2=4+
%28y-5%29%5E2=0
y-5=0
y=5
=> y- intercept is at (0,5)
since x+is squared, there must be one more point (x,5), so
substitute y=5 in %28x%2B2%29%5E2+%2B+%28y-5%29%5E2=4.....eq.1 and solve for x+to get one more point on the circle:
%28x%2B2%29%5E2+%2B+%285-5%29%5E2=4+
%28x%2B2%29%5E2+=4+
x%2B2=sqrt%284+%29
x%2B2=2 or x%2B2=-2
x=0 or x=-4
solutions:
x=0, y=5+=> point (0,5)
x=-4, y=5=> second point (-4,5)

since solutions should be same for both eq.1 and eq.2, then go to
-4x%2By=k......eq.2, substitute x+and y values from points (-2,5) , (0,5) and (-4,5)
-4%2A-2%2B5=k
k=13

-4%2A0%2B5=k
k=5

-4%2A-4%2B5=k
k=21
so, maximum value of k is 21

Answer by ikleyn(52818) About Me  (Show Source):
You can put this solution on YOUR website!
.

In this my post,  I want to notice that the problem is posed incorrectly.

Indeed, it says that  x, y ∈ Z,  which is  INCORRECT.

A circle with the center at rational point on a plane (in this problem - a circle with the center
at integer point  (-2,5))  and the radius  2  can not contain any rational/integer point,  except trivial points
((0,5),  (-2,7),  (-4,5)  and  (-2,3).

The correct form is   x, y ∈ R.


Also,  keep in mind that the solution by @MathLover1 is  INCORRECT  and  IRRELEVANT.


/////////////////////


Edwin, it is somehow awkward for me to read and to comment what you wright on this your post as mccravyedwin.




Answer by mccravyedwin(408) About Me  (Show Source):
You can put this solution on YOUR website!
Ikleyn is apparently not familiar with the standard 
notation of Z for the set of all integers.  I had
forgotten it myself and did not notice it. But the
solution by MathLover1 is the correct one. Mine at
the top is correct for the actual maximum value of
k but required some "eyeballing" to find the actual
maximum, assuming x and y to be integers.  Copied
from the internet:

The set of integers is represented by the letter Z. 
An integer is any number in the infinite set,

Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}

Integers are sometimes split into 3 subsets, 

Z%5E%22%2B%22, Z%5E%22-%22 and {0}. Z%5E%22%2B%22 is the 
set of all positive integers (1, 2, 3, ...), while Z%5E%22-%22
is the set of all negative integers (..., -3, -2, -1). 
Zero is not included in either of these sets . Z%5Enonneg 
is the set of all positive integers and 0, while Z%5Enonpos 
is the set of all negative integers and 0. 

Edwin