+9479 Can you apply the same method to solve your actual question?
First find the x-intercepts of each graph. (Plug in 0 for y and solve for x )
Then find the points of intersection. (The solutions to \(\frac{(x-k)^2}{9}+y^2\ =\ \frac{x^2}{9}+y^2\) are the points of intersection)
Once you know the coordinates of each point A, B, C, and D, then find the slope of AB and the slope of BC.
(If you calculate the slope of CD and AD you should find that they are the same as AB and BC, respectively)
Here's a new graph: https://www.desmos.com/calculator/x63mqeac9l
Then find which value of k will make the slope of BC = the negative reciprocal of the slope of AB
If you need more help or get stuck let us know!
+9479 I don't know how to prove it, but your question inspired me to create this regular polygon grapher on desmos:
https://www.desmos.com/calculator/br9gidymwe
I spent all afternoon working on it, and I thought it was really cool, so I wanted to share 
You can select the number of sides with the slider for n , and go into the drawings folder to turn off/on certain parts of the graph
( Also another cool use for this graph is to create diagrams for questions 
)
+9479 The roots of the quadratic equation \(x^2+bx+c=0\) are:
\(x=\frac{-b+\sqrt{b^2-4c}}{2}\) and \(x=\frac{-b-\sqrt{b^2-4c}}{2}\)
The difference between these two roots is:
\(\frac{-b\ +\ \sqrt{b^2-4c}}{2}-\frac{-b\ -\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{-b\ +\ \sqrt{b^2-4c}}{2}+\frac{b\ +\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{-b\ +\ \sqrt{b^2-4c}\ +\ b\ +\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{ \sqrt{b^2-4c}\ +\ \sqrt{b^2-4c}}{2}\\~\\ =\quad\frac{ 2\sqrt{b^2-4c}}{2}\\~\\ =\quad\sqrt{b^2-4c}\)
which we are told must be equal to \(|b-2c|\) ...so we can make this equation:
\(\sqrt{b^2-4c}\ =\ |b-2c|\) Now let's solve this equation for c
\((\sqrt{b^2-4c}\ )^2\ =\ (\ |b-2c|\ )^2\)
Squaring a negative number gives the same result as squaring the positive version of that number,
so we can drop the absolute value signs
\(b^2-4c\ =\ (b-2c)^2 \\~\\ b^2-4c\ =\ (b-2c)(b-2c) \\~\\ b^2-4c\ =\ b^2-4bc+4c^2 \)
Subtract b2 from both sides of the equation
\(-4c\ =\ -4bc+4c^2 \)
Add 4c to both sides of the equation
\(0 =\ -4bc+4c^2 + 4c\)
Rearrange the terms
\( 0 =\ 4c^2 -4bc+ 4c \)
Divide through by 4
\( 0 =\ c^2 -bc+ c\)
Factor c out of all three terms on the right sidde
\( 0 =\ c(c -b+ 1)\)
Set each factor equal to zero and solve for c
\(\begin{array}{ccc} c=0&\quad\text{or}\quad&c-b+1=0\\ &&c-b=-1\\ &&c=b-1 \end{array}\)
Check: https://www.wolframalpha.com/input/. . .
So either c = 0 or c = b - 1
We are given that c ≠ 0,
So it must be that c = b - 1
+9479 Did you mean to put \(\frac{(x - k)^2}{9} + y^2 = 1\) and \(\frac{(x + k)^2}{9} + y^2 = 1\) ? If so.....
By finding the x-intercepts and the points of intersection (in terms of k), we can find the slope of line AB, BC, CD, and AD.
Then, once we know a point and the slope of each line, we can find the equations of each line.
And so we can make this graph: https://www.desmos.com/calculator/uozsru2uca
Now we need to find the value of k that makes AB perpendicular to BC
(which will also make AB perpendicular to AD, and CD perpendicular to BC, and CD perpendicular to AD)
That is, we need to find the value of k that makes the following true:
the slope of BC = the negative reciprocal of the slope of AB
\(-\left(\frac{\sqrt{9-k^{2}}}{9-3k}\right)\quad=\quad\)the negative reciprocal of \(\left(\frac{\sqrt{9-k^{2}}}{9-3k}\right)\)
\(-\left(\frac{\sqrt{9-k^{2}}}{9-3k}\right)\quad=\quad-\left(\frac{9-3k}{\sqrt{9-k^{2}}}\right)\)
Multiply both sides of the equation by -1
\(\frac{\sqrt{9-k^{2}}}{9-3k}\quad=\quad\frac{9-3k}{\sqrt{9-k^{2}}}\)
Multiply both sides by \(\sqrt{9-k^2}\) and multiply both sides by \(9-3k\)
\((\sqrt{9-k^2}\ )^2\quad=\quad(9-3k)^2\)
Simplify the left side and multiply out the right side
\(9-k^2\quad=\quad(9-3k)(9-3k)\)
\(9-k^2\quad=\quad81-54k+9k^2\)
Subtract 9 from both sides and add k2 to both sides
\(0\quad=\quad72-54k+10k^2\)
Rearrange the terms
\(0\quad=\quad10k^2-54k+72\)
Now we can use the quadratic formula to find k
\(k\quad=\quad\dfrac{54\pm\sqrt{54^2-4\cdot10\cdot72}}{2\cdot10}\)
\(k\quad=\quad\dfrac{54\pm6}{20}\)
\(k\quad=\quad3\qquad\text{or}\qquad k\quad=\quad\frac{12}{5}\)
If k = 3 then the points A, B, C, and D are all on the origin. But if k = 12/5 then ABCD is a regular square.
If you would like more explanation on how to find the slopes of lines BC and AB then please let us know!
