Find the equation of the locus of a point so that the square of its distance from (3, -3) is always numerically equal to the slope of the line joining it to the same point.
"\\displaystyle\n\\text{Consider the point(x,y):}\\\\\np^2((x,y);(3,-3)) = (x-3)^2 + (y+3)^2\\\\\n\\text{The slope of corresponding line equals $\\frac{y+3}{x-3}$. As a result, we have an equation:}\\\\\n(x-3)^2 + (y-3)^2=\\frac{y+3}{x-3}"
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