How is this inequality plotting a circle? $3\sqrt{x^2+y^2}\leq 10-x^2-y^2$

Question

I'm trying to understand how exactly this inequality represents the inner part of a circle:

$$3\sqrt{x^2+y^2}\leq 10-x^2-y^2$$

I tried taking the squares of both members but what I get in the end is this

$$-(x^4+y^4)+ 29x^2+29y^2 -2x^2y^2 \le 100$$

Which doesn't represent to me a circle, which is what WolframAlpha would plot to me when I plug in the first one

What am I missing?

Answer 1

Let $z=\sqrt{x^2+y^2}>0$ then

$$3\sqrt{(x^2+y^2)}\le 10-x^2-y^2 \iff 3z\le10-z^2 $$$$\iff z^2+3z-10 \le 0 \iff (z+5)(z-2)\le0$$

and since $z+5 >0$ the condition corresponds to

$$z-2 \le 0 \iff x^2+y^2\le 4$$

Answer 2

Because it's $$\left(\sqrt{x^2+y^2}-2\right)\left(\sqrt{x^2+y^2}+5\right)\leq0$$ or $$x^2+y^2\leq4.$$

Answer 3

Note that your equation may be rewritten entirely in terms of $r=\sqrt{x^2+y^2}$, the distance from the origin to the point $(x,y)$. Since the only thing about a point which affects whether it satisfies the inequality is its distance to the origin, the resulting region must be rotationally symmetric.

It isn't hard to see for which values of $r$ the inequality holds, and that it gives a single disc (rather than, say, an annulus or something).

Answer 4

Making the change of variables

$$ x = r\cos\theta\\ y = r\sin\theta $$

we have

$$ 3r \le 10-r^2 $$

solving for $r$ we have

$$ 0 \le r \le 2 $$

thus representing in this range the circle's interior plus boundary.

Answer 5

When you square you are losing some information, given that the right hand side can be negative while the left hand side cannot. If you want to square you have to impose that the RHS has to be positive (if the RHS is negative than you have no solution), i.e. $$x^2+y^2\le10$$ Squaring $$ x^4+2x^2\,y^2+y^4-29 x^2 -29 y^2+100\ge 0$$ $$ (x^2+y^2)^2-29 (x^2 + y^2)+100\ge 0$$ $$ (x^2+y^2-25)(x^2+y^2-4)\ge 0$$ that is verified when the factors have same sign, i. e $$x^2+y^2\le25 \cap x^2+y^2\le4$$ $$\implies x^2+y^2\le4 $$ and $$x^2+y^2\ge25 \cap x^2+y^2\ge4$$ $$\implies x^2+y^2\ge 25 $$ Intersecting with the previous condition $$x^2+y^2\le4 \cap x^2+y^2\le10 $$ $$\implies x^2+y^2\le4 $$ and $$x^2+y^2\ge25 \cap x^2+y^2\le10 $$ that has no solution. So the only solution is $$x^2+y^2\le4 $$