Difficult problem of Maxima and minima

Question

Let $x$ and $y$ be real numbers $$ x^2 + y^2 = 2x – 2y + 2.$$ What is the largest possible value of $$x^2 + y^2 – \sqrt{32}\; ?$$

I tried the problem by rearranging the equation as $x^2 -2x -2 = - y^2 - 2y$ As coefficient of $y^2 < 0$, maximum value of $x^2 -2x -2 = 1$. [$(Y)max=-D/4a$]

Now putting $x^2 -2x - 3 = 0$, $x$ turns out to be $3$ or $-1$. As $x^2 -2x -2 = - y^2 - 2y$ and $x^2 -2x -2=1, - y^2 - 2y=1$. On solving, $y = -1$

Now putting it in $x^2 + y^2 - \sqrt{32}$, the answer turns out to be $10 - \sqrt{32}$. But the answer is $6$. How??

Answer 1

The problem is not so difficult.

The equation $x^2+y^2=2x-2y+2$ can be rewritten as $$(x-1)^2+(y+1)^2=4$$ which is an equation of a circle centred at $(1,-1)$ with radius $2$. The maximum of $x^2+y^2-\sqrt{32}$ on this circle is clearly attained where the maximum of $x^2+y^2$ is attained on the same circle, so we are looking for a point on our circle whose (square of the) distance to the origin is maximal. Elementary geometry reveals that such a point must lie on the line passing through $(0,0)$ and the centre of the circle, $(1,-1)$. The equation of this line is $y=-x$, and it intersects our circle at two points: $(1-\sqrt{2},\sqrt{2}-1)$ and $(1+\sqrt{2},-1-\sqrt{2})$. The second point is the farthest one, and the distance squared is $2(1+\sqrt{2})^2$. Therefore, the original functions' maximum over the circle is equal to $$2(1+\sqrt{2})^2-\sqrt{32}=2(1+2\sqrt{2}+2)-4\sqrt{2}=6$$

Answer 2

WLOG $x=2\cos t+1, y=2\sin t-1$

$$x^2+y^2=(2\cos t+1)^2+(2\sin t-1)^2=2+1+1+4(\cos t-\sin t)=4+4\sqrt2\cos\left(t+\dfrac\pi4\right)$$

Now as $t$ is real, $$-1\le\cos\left(t+\dfrac\pi4\right)\le1$$