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.
Consider the point(x,y):p2((x,y);(3,−3))=(x−3)2+(y+3)2The slope of corresponding line equals y+3x−3. As a result, we have an equation:(x−3)2+(y−3)2=y+3x−3\displaystyle \text{Consider the point(x,y):}\\ p^2((x,y);(3,-3)) = (x-3)^2 + (y+3)^2\\ \text{The slope of corresponding line equals $\frac{y+3}{x-3}$. As a result, we have an equation:}\\ (x-3)^2 + (y-3)^2=\frac{y+3}{x-3}Consider the point(x,y):p2((x,y);(3,−3))=(x−3)2+(y+3)2The slope of corresponding line equals x−3y+3. As a result, we have an equation:(x−3)2+(y−3)2=x−3y+3
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