# To find the locus of a point

A point moves such that the sum of its distances from the coordinate axes is equal to its distance from the circle $x^2+y^2=4$.

I tried and got this equation for the locus: $$x+y=\sqrt{x^2+y^2}-2$$

Should I replace $x+y$ with $|x|+|y|$ because distances are absolute? In that case, how will I represent $y$ as an explicit function of $x$?

Not only should you replace $x$ and $y$ with $|x|$ and $|y|$, you should also take the absolute value of the right-hand side because it represents a distance too: $$|x|+|y|=\left\vert\sqrt{x^2+y^2}-2\right\vert$$ To solve this for $y$, we can first restrict ourselves to non-negative $x$ and $y$, since substituting $x\to-x$ or $y\to-y$ in a valid solution yields another valid solution. With this restriction we can remove the modulus signs around $x$ and $y$. The triangle inequality implies that $x+y>\sqrt{x^2+y^2}-2$, so when removing the modulus signs in the RHS we must have $$x+y=2-\sqrt{x^2+y^2}$$ which leads to $$2-(x+y)=\sqrt{x^2+y^2}$$ $$4-4(x+y)+x^2+2xy+y^2=x^2+y^2$$ $$4-4(x+y)+2xy=0$$ $$2-2(x+y)+xy=0$$ $$(x-2)y+2(1-x)=0$$ $$y=\frac{2(x-1)}{x-2},x\le1$$ Mirroring this across the $x$- and $y$-axes we get the final equation for $y$ in terms of $x$: $$y=\pm\frac{2(|x|-1)}{|x|-2},|x|\le1$$ Here is a plot of the locus asked for by the question, with $x^2+y^2=4$ added as a reference.

• great solution. Can you just mention from where you got the constraint x<1 Oct 6 '16 at 11:17
• @AmritanshSinghal The constraint is to eliminate an unwanted solution that occurs when I square both sides of $2-(x+y)=\sqrt{x^2+y^2}$. Oct 6 '16 at 11:19

Let the point of the locus be $(a,b)$.

It's equivalent to two circles are touching

$$\left \{ \begin{array}{rcl} (x-a)^2+(y-b)^2 &=& (|a|+|b|)^2 \\ x^2+y^2 &=& 4 \end{array} \right.$$

$$\left \{ \begin{array}{rcl} x^2+y^2-2ax-2by &=& 2|a b| \\ x^2+y^2 &=& 4 \end{array} \right.$$

Eliminating quadratic terms: \begin{align*} ax+by &=2-|ab| \\ y &= \frac{2-|ab|-ax}{b} \\ x^2+\left( \frac{2-|ab|-ax}{b} \right)^2 &=4 \\ b^2x^2+[a^2x^2-2a(2-|ab|)x+(2-|ab|)^2] &= 4b^2 \\ (a^2+b^2)x^2-2a(2-|ab|)x+(2-|ab|)^2-4b^2 &=0 \\ \Delta &= 0 \\ a^2(2-|ab|)^2-(a^2+b^2)[(2-|ab|)^2-4b^2] &= 0 \\ 4b^2(a^2+b^2)-b^2(2-|ab|)^2 &= 0 \\ 4(a^2+b^2)-(2-|ab|)^2 &= 0 \\ 4(a^2+b^2)-a^2b^2+4|ab|-4 &= 0 \\ \end{align*}

Note that distance from circle is either maximum or minimum.

Considering the minimal distance only:

If maximal distance is also included:

• can you please elaborate a little on how you got the equivalent condition ? Oct 6 '16 at 10:29
• If two circles make two intersections, the point $(a,b)$ is too close to the cirlce $x^2+y^2=4$; whereas if no intersections, the point is too far apart. Tangency is just right. Oct 6 '16 at 13:55