If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$

Question

Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that $$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$

I'm struggling with that problem. I've recognized that the given equation is one of a circle, namely; $$(x-2)^2 + (y+5)^2 = 9$$

And what needs to be proven can be derived to the form: $$7-3\sqrt{2}\le x-y\le 7 + 3\sqrt{2} \quad\Leftrightarrow\quad (x-y)^2 - 14(x-y) + 31 \le 0$$

But I really can't find the correlation between what's given and what has to be proven! Also, I'm wondering if this problem can be generalized for any circle parameters, something like an universal interval of $x-y$ for any circle. Any help is apprecatiated! Thank you in advance!

Answer 1

Hint:

Let $x-y=c$

Use $x=y+c$ to form a Quadratic Equation in $y$

As $y$ is real, the discriminant must be $\ge0$

One may replace $y$ with $x-c$ as well.

Answer 2

WLOG $x=2+3\cos t, y=-5+3\sin t$

$$x-y=7+3(\cos t-\sin t)$$

Now $$(\cos t-\sin t)^2+(\cos t+\sin t)^2=2$$

$$\implies(\cos t-\sin t)^2\le2$$

Answer 3

Think about it geometrically; $x-y = c$ is a line parallel to $y=x$, and you want to find all such $c$ that makes the line intersect the circle.

This makes it clear that you only need to find the two lines that are tangent, and the tangency points are at $(2,-5) \pm (\frac{3}{\sqrt 2}, -\frac{3}{\sqrt 2})$ (just draw two lines tangent and you'll see how this is clear). Compute the values of $x-y$, and your answer will appear.

Answer 4

Writing $u = x - 2$ and $v = y + 5$ shows that the problem is equivalent to proving that if $u^2 + v^2 = 9,$ then $|u - v| \leqslant 3\sqrt2.$ We have $$ (u - v)^2 + (u + v)^2 = (u^2 - 2uv + v^2) + (u^2 + 2uv + v^2) = 2(u^2 + v^2) = 18, $$ therefore $(u - v)^2 \leqslant 18,$ and the result follows by taking square roots.

More generally, for the circle $(x - a)^2 + (y - b)^2 = r^2,$ if we write $u = x - a$ and $v = y - b,$ essentially the same argument gives $(u - v)^2 \leqslant 2r^2,$ whence $|u - v| \leqslant r\sqrt2,$ i.e. $$ a - b -r\sqrt2 \leqslant x - y \leqslant a - b + r\sqrt2. $$