How would you solve these equations and show that they do not intersect each other?
$$x^2+y^2=2x-2y$$
$$x^2+y^2=4(x^2+y^2)^{1/2} +y$$
It's isolating a term which I am struggling with.
General guidance or even a partial solution would be much appreciated.
On
Since you have solved both for $x^2+y^2$, set the two equations equal to each: $$2x-2y=4(x^2+y^2)^{1/2}+y$$
Now simplify as much as possible to obtain $$12x^2+7y^2+12xy=0$$
One method to see that the curves do not intersect is to solve for one of the variables and look at the discriminant. Solving for $y$: $$y=\frac{-12x\pm\sqrt{(12x)^2-4(7)(12x^2)}}{2(7)}$$
Now look at the discriminant (expression beneath the radical): once simplified, it results in $-192x^2$. Assuming $x\ne0$, $x^2$ is always a positive value and thus $-192x^2$ is negative. Therefore there are no real solutions to the equation, and thus the original two curves do not intersect, except at $x=0$.
On
How would you solve these equations and show that they do not intersect each other?
I create a plot to get a first overview and then decide how to proceed.
These look like circles and would make bring the equations into the form $$ (x - cx)^2 + (y - cy)^2 = r^2 $$ to be able to read center and radius. And from this argue that the circles do not intersect.
However, looking closer, the second curve (green colour) seems to be no circle albeit something very close to a circle, so it gets more complicated.
First curve (red colour): $$ x^2 + y^2 = 2x - 2y \iff \\ (x-1)^2 + (y + 1)^2 = 1 + 1 = (\sqrt{2})^2 $$
Second curve (green colour): $$ x^2+y^2=4(x^2+y^2)^{1/2} +y $$
In polar coordinates, the equation turns into $$ r^2 = 4 r + r \sin \phi $$ and this means $$ r^2 \ge 4r - r = 3r \iff r \ge 3 \vee r = 0 $$ For the first curve we have $$ r^2 = 2r (\cos \phi - \sin \phi) = 2 r \sqrt{2} \sin(\pi/4 - \phi) \le 2 \sqrt{2} r \iff \\ r \le 2 \sqrt{2} < 3 \vee r = 0 $$ because $(2\sqrt{2})^2 = 8 < 9 = 3^2$. So $(0,0)$ is a common solution to both curves.
Otherwise if $C$ is the circle around the origin with radius $3$ then the first curve lays within $C$ and the second curve lays no closer than on $C$, they are separated by $C$. So there are no more common solutions.
On
The answers you have received don't not solve separately each equation as you ask so I do it. $$(1)\space\space x^2+y^2=2x-2y\iff(x-1)^2+(y+1)^2=2$$ it is clear there are infinitely many real solutions $(x,y)$ in the corresponding circle. You have in particular the integers ones $(0,-2),(0,0),(2,-2)$.
$$(2)\space\space x^2+y^2=4(x^2+y^2)^{1/2} +y\iff (x^2+y^2-y)^2-16(x^2+y^2)=0$$ Similarly,infinitely many real solutions in the graph of the relation which seems to be a circle (look at the figure above). You have in particular the integer solutions $(\pm4,0),(0,0),(0,5),(0,-3)$.
►Looking at the circle $(1)$ and what seems to be one $(2)$ it is obviuos that the curves does not intersect (but it was not my intention to answer this second part of the question).
On
When converted to polar coordinates there is an advantage as seen here. The first curve is a cardoid
$$ r = \sin \theta +4 $$
The maximum/minmum distances from origin are $5,3.$
The second one is a circle of radius $ =\sqrt 2 ,$ tangent to a straight line $ \theta = \pi/4$ passing through origin.
They do not cut anywhere as the circle diameter
$$ 2 \sqrt 2 < 3$$
with a minimum distance of
$$ 3- 2 \sqrt 2 $$
between the circle and cardoid.
Pass to polar coordinates: $x=r\cos\varphi$ and $y=r\sin\varphi$. The equations become $$ r^2=2r(\cos\varphi-\sin\varphi), \qquad r^2=4r+r\sin\varphi $$ Excluding the solution $r=0$, we get $$ \sin\varphi=r-4, \qquad \cos\varphi=\frac{3}{2}r-2 $$ Since $\sin^2\varphi+\cos^2\varphi=1$, we must have $$ r^2-8r+16+\frac{9}{4}r^2-6r+4=1 $$ or $$ 13r^2-56r+76=0 $$ that has no solution.