find an equation in x and y such that the distance between (x,y) and (-2,0) is twice the distance between (x,y) and (3,1)

Answer 1

I assume you want to find a point (x, y) that fits the conditions.

Find the distance between (-2,0) and (3,1) but keep it in x,y form.

(3-(-2),1-0) = (5,1)

It may be helpful now for you to draw a graph to see whether to add (5,1) to (3,1) or subtract (5,1) from (-2,0).

However, because the distance between (x,y) and (-2,0) is twice the distance between (x,y) and (3,1), you add it to (3,1). (This is why a graph is helpful.)

So, add the distance (5,1) to (3,1).

(3+5, 1+1) = (8,2)

Answer 2

Translating your condition gives me

√((x+2)^2 + y^2) = 2√(x-3)^2 + (y-1)^2)
square both sides
(x+2)^2 + y^2 = 4((x-3)^2 + (y-1)^2)
x^2 + 4x + 4 + y^2 = 4(x^2 - 6x + 9) + y^2 - 2y + 1)
expanding, collecting like terms and simplifying I got

3x^2 + 3y^2 + 20x - 8y = -36

which has the form of a circle, so I completed the square to end up with

(x + 10/3)^2 + (y - 4/3)^2 = 224/9