$M_1 = (x,y)\quad x²+y²+6y = 7 $ to $x \rightarrow y$

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I have two relations:

$$M_1 = (x,y)\qquad x²+y²+6y = 7 $$

$$M_2 = (x,y)\qquad x²+y²-6x = 7, \qquad y \ge 0$$

The question is if this relations also reflex functions like $x \rightarrow y$?

I think I have to test if I also can write the relations like $x \rightarrow y$ right?

So what I did was:

$$M_1 = (x,y)\qquad x^2+y^2+6y = 7 $$

$$M_1 = (x,y)\qquad x= \sqrt{-y^2-6y+7} $$

My question us can i short it better? And is $M_1$ now a function like $x \rightarrow y$? I would say that like i have it now it wouldnt be the same like $x \rightarrow y$ because of the $y^2$?

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$$M_1:x^2+y^2+6y=7\iff x^2+y^2+2\cdot3y+3^2=7+3^2\iff x^2+(y+3)^2=16=4^2$$

$\iff M_1$ is a circle centered in $(0,-3)$ with radius $4$. Also, $y(x)=-3\pm\sqrt{16-x^2}$.


$$M_2:x^2+y^2-6y=7\iff x^2+y^2-2\cdot3y+3^2=7+3^2\iff x^2+(y-3)^2=16=4^2$$

$\iff M_2$ is a circle centered in $(0,3)$ with radius $4$. Also, $y(x)=3\pm\sqrt{16-x^2}$.