Find the equation of the locus. Help please?

Let the given point be A(3, -3) and let P(x, y) be a point on the locus.

(PA)² = (x - 3)² + (y + 3)²

slope of PA = (y + 3)/(x - 3), for x ≠ -3

(x - 3)² + (y + 3)² = (y + 3)/(x - 3)

(x - 3)³ + (x - 3)(y + 3)² - y - 3 = 0

Let be a point M(x,y).

Distance from M(x,y) to point(3,-3) is

D^2 (x,y) = (x-2)^2 + (y-3)^2

Slope of line is passing to M(x,y) and (3,-3) is (-3-y)/(3-x)

Then you get

(x-2)^2 + (y-3)^2 = (-3-y) / (3-x) THis is the equation of the locus :

" the square of its distance from (3,-3) is always numerically equal to the slope of the line joining it to the same point"

square of distance from(x, y) to (3,-3) = slope of the line joining (x, y) to (3, -3)

(x - 3)^2 + (y + 3)^2 = (y + 3) / (x - 3)

(x - 3)^3 + (x -3)(y + 3)^2 - (y + 3) = 0

x^3 - 9 x^2 + x y^2 + 6 x y + 36 x - 3 y^2 - 19 y - 57 = 0