I'm aware that, if I have two functions defined with z as the independent variable, like $z=x^2+y^2$ and $z=x+2y+5$ I can just set them as equal to each other: $$x^2+y^2=x+2y+5$$ $$x^2+y^2-x-2y=5$$ $$(x-\frac{1}{2})^2+(y-1)^2=\frac{15}{4}$$ I don't recognize what kind of curve that is, but I guess it's the intersection of the first two, is that right?
Well, now what happens when I have something like $z=x+y$ and $x^2+y^2=4$? What other methods can I use to find an intersection between curves like these?
In the first case, you have a circle, centered at $(\frac{1}{2}, 1)$.
To answer your second question, you always use the same method: you find out what you can say about the variables if you assume both equations are true.
From the second equation, you have $$x^2=4-y^2$$
Substituting into the second equation, we get the intersection $$z = y+\sqrt{4-y^2}$$ And $$z = y-\sqrt{4-y^2}$$ Any point on this curve will satisfy both equations.
You can also put both equations in polar coordinates, then you will get a parametrization in terms of $\theta$. The fact that $$\cos^2+\sin^2=1$$ makes this much simpler.