Find the locus of a point such that the square of its distance to the point of intersection of two perpendicular lines is equal to the sum of its distances to those lines.
Assume $P(x,y)$ is any point of the locus, then:
$$x^2+y^2=\sqrt{x^2+(y-2)^2}+\sqrt{(x-2)^2+y^2}$$
Am I doing it right? (intersection point at $(0,0)$, points A and B at $(0,2)$ and $(2,0)$.)
Best Answer
Take the x-axis and the y-axis as the two perpendicular lines, so that the intersection is the origin. The square of the distance to the origin is $x^2+y^2$. The sum of the distances to the axes is $|x|+|y|$. So the locus is
$$x^2+y^2=|x|+|y|$$
Note that this function is symmetric for both $x$ and $y$. So, to draw it, it is sufficient to solve it in the first quadrant and then draw its symmetric curves in the other quadrants. In the first quadrant, it corresponds to a semicircle with center in $(0.5,0.5)$ and radius $\displaystyle \frac{\sqrt{2}}{2}$ that intersects the axes in the points $(0,1)$ and $(1,0)$. Tracing the symmetric curves in the other quadrants, we get that the whole locus has the shape of a four-leaved clover with semicircular leaves.