Find an equation for the set of all points P(x, y) where its distance from (2, 5) is twice its distance from (-3, 1). Please explain :)?

Put A = (2,5) and B = (-3,1) and P = (x,y). Now AP = 2BP, ie.,
4(BP)^2 - (AP)^2 = 0, ie., 4[(x+3)^2 + (y-1)^2] - [(x-2)^2 + (y-5)^2] = 0,
ie., (2x+6)^2 + (2y-2)^2 - (x-2)^2 - (y-5)^2 = 0, ie., (2x+6-x+2)(2x+6+x-2)
+ (2y-2-y+5)(2y-2+y-5) = 0, ie., (x+8)(3x+4) + (y+3)(3y-7) = 0, ie.,
3x^2 +28x +32 +3y^2 +2y -21=0, ie., x^2+(28/3)x +y^2+(2/3)y = -(11/3),
ie., (x+[14/3])^2 + (y+[1/3])^2 = (197/9) - (33/9) = [(2/3)rt41]^2. This is
the eqn of a circle, center (-[14/3],-[1/3]), radius (2/3)rt41.

Distance between P(x, y) and (2, 5)

= sqtr[ (x - 2)^2 + (y - 5)^2

Distance between P(x, y) and (-3, 1)

= sqrt[ (x + 3)^2 + (y - 1)^2

Therefore

sqtr[ (x - 2)^2 + (y - 5)^2 = 2sqrt[ (x + 3)^2 + (y - 1)^2

Squaring blth sides

(x - 2)^2 + (y - 5)^2 = 4[ (x + 3)^2 + (y - 1)^2]

x^2 - 4x + 4 + y^2 - 10y + 25 = 4[ x^2 + 6x + 9 + y^2 - 2y + 1]

x^2 + y^2 - 4x - 10y + 29 = 4x^2 + 4y^2 + 24x - 8y + 40

3x^2 + 3y^2 + 28x + 2y + 11 = 0

Your equation is
(x-2)²+(y-5)²=4(x+3)²+4(y-1)²
or
3x²+28x+3y²+2y+11=0
or
(x+14/3)²+(y+1/3)²=164/9
Thus, the set is a circle of radius r=√164/3 centered at a point (-14/3, -1/3).