Let $x,y\in\Bbb R$ such that $x ^2 + y^2 = 2x – 2y + 2$. What is the largest possible value of $x^2 + y^2 -\sqrt{32}$??

Question

I tried to make perfect squares by adding 2 on both sides and I got something which looked like a Pythagorean Triplet but I am stuck there. Also is there a well defined way or an algorithm which gives me the least/ maximum positive value of the solutions of a diophantine equation (specifically a linear diophantine equation.) Thanks!

Answer 1

You have

$$\begin{equation}\begin{aligned} x^2 + y^2 & = 2x - 2y + 2 \\ x^2 - 2x + 1 + y^2 + 2y + 1 & = 2 + 2 \\ (x - 1)^2 + (y + 1)^2 & = 4 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

I believe this equation is what basically what you were referring to in your initial statement of

I tried to make perfect squares by adding 2 on both sides and i got something which looked like a Pythagorean Triplet but i am stuck there.

This forms a circle with center $(1,-1)$ and radius $2$, as Rezha Adrian Tanuharja's question comment indicates, and as shown in the diagram below.

The value of $x^2 + y^2$ is the square of the distance of a point on this circle from the origin, shown as $A$ above. The line $AOB$ has the point $B(1 + \sqrt{2}, -1 - \sqrt{2})$. This is the point which gives the largest distance because, using any other point on the circle, such as $C$, with $\triangle AOC$, you have $|OC| = |OB| = 2$, so the triangle inequality states

$$|AB| = |AO| + |OB| = |AO| + |OC| \gt |AC| \tag{2}\label{eq2A}$$

Note that, similarly, the minimum distance would be the point on the other end of the diameter. As such, you get a maximum value of

$$\begin{equation}\begin{aligned} x^2 + y^2 - \sqrt{32} & = (1 + \sqrt{2})^2 + (-1 - \sqrt{2})^2 - 4\sqrt{2} \\ & = (1 + 2\sqrt{2} + 2) + (1 + 2\sqrt{2} + 2) - 4\sqrt{2} \\ & = 6 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Answer 2

Hint:

First it is equivalent to determining the largest value of $x^2+y^2$, i.e. of the distance from the origin (squared).

Then do some geometry: as mentioned in the comment, a point $M(x,y)$ satisfying the relation belongs to the circle with radius $2$ centred at $(1,-1)$. The point on this circle which is farthest from the origin is on the diameter though the origin, i.e. on the line with equation $x+y=0$.

Answer 3

You can calculate the maximum directly.

Rewriting the restriction as $$(x-1)^2+(y+1)^2=4$$ you can set

$$x=1+2\cos t, y= -1+2\sin t, t\in [0,2\pi]$$

Plugging this into $x^2+y^2$ you get $$x^2+y^2=2+4+4\sqrt{2}\cos\left(t+\frac{\pi}4\right)\leq 6+\sqrt{32}$$

Hence, the searched for maximum is $\boxed{6}$.

Answer 4

A circle: Centre $(1,-1);$ $r=2$

$(x-1)^2+(y-(-1))^2=2^2$;

Symmetry about line: $y=-x$ (Centre of circle $(1,-1)$ lies on $y=-x$).

Intersection of line $y=-x$ with circle:

$(x-1)^2+(-x+1)^2=4.$

$2(x-1)^2=4$; $(x-1)^2=2$;

$x_{1,2}=1\pm√2$; $y_{1,2}=-1 \mp √2.$

Largest distance$^2$:

$x_1^2+y_1^2=$

$(1+√2)^2+(-1-√2)^2=$

$2(1+2√2+2)=6+4√2$;

Can you finish?

Answer 5

I was seeing the answers to the question , all were using coordinate geometry or other complex ways to solve the equation , but there is another simple way.

Just transpose the equation to make it $x^2-2x+y^2-2y=2$

Now if we see we can just put $x^2-2x$ and $y^2-2y$ equal to 1 as $1+1=2$ From the quadratic formula we get

$x=1+\sqrt2$ or $x=1-\sqrt2$ $y=1+\sqrt2$ or $y=1-\sqrt2$

Now to get max value of $x^2+y^2-\sqrt32$ we choose $1+\sqrt2$ as value of x and y

So we get $x^2+y^2-\sqrt32$ as 6

I think this is a much elegant method than the above ones.