The graph of $\sqrt{(x+2)^2+y^2} + \sqrt{x^2 + (y-2)^2} = 6$ is a(n)...
a. Ellipse; b. Hyperbola; c. Parabola; D. circle; E.Straight line.
I originally wanted to square both sides. But seeing how that would quickly get messy, and how the two terms are almost similar, I did $\sqrt{(x+2)^2+y^2} = 3$ and $\sqrt{x^2 + (y-2)^2} = 3$ and squared both sides. Adding the equations and simplifying, I got $(x+1)^2 + (y-1)^2 = 7$. But I'm not even sure this is the right way, and I think there should be a better approach... Any help would be appreciated!
The Sum of distance from two points is constant. The locus is an ellipse. The foci are $S_1(-2,0)$ and $S_2(0,2)$. Also we check $S_1S_2 = 2 \sqrt{2} <6$ so indeed its an ellipse.
$$PS_1 + PS_2 = 6$$
This check $S_1S_2 < 6$ is necessary because otherwise it will be empty set (ref. Triangle inequality).