Given locus is a circle, prove two lines are perpendicular

Question

Let $l_1$ and $l_2$ be two lines in the plane. The locus of all points $P$, such that the sum of squares of the distances of $P$ to $l_1$ and $l_2$ is constant, is a circle. Prove that $l_1$ and $l_2$ are perpendicular.

Now I can prove the converse of this statement really easily, but I'm stuck on proving this. I've let the centre of the circle be (0,0); I'm not sure if this helps. Also, how can we assume that the two lines will intersect at the centre of the circle?

Answer 1

Let $l_1$ be given by $a_1x+b_1y+c_1=0$ and $l_1$ be given by $a_2x+b_2y+c_2=0$.

Let $P$ be the point $(h,k)$ and $d_j$ be the distance of the point $P$ from line $l_j$. Then, $$d_1=\frac{|a_1h+b_1k+c_1|}{\sqrt{a_1^2+b_1^2}} \quad \text{ and } \quad d_2=\frac{|a_2h+b_2k+c_2|}{\sqrt{a_2^2+b_2^2}}$$

We are given that the locus of the points for which $d_1^2+d_2^2=s$ (where $s$ is some constant) is a circle. Let $\lambda_1=a_1^2+b_1^2$ and $\lambda_2=a_2^2+b_2^2$. Observe that $$d_1^2+d_2^2=\left(\frac{\lambda_2a_1^2+\lambda_1a_2^2}{\lambda_1 \lambda_2}\right)h^2+\left(\frac{\lambda_2b_1^2+\lambda_1b_2^2}{\lambda_1 \lambda_2}\right)k^2+\left(\frac{2a_1b_1}{\lambda_1}+\frac{2a_2b_2}{\lambda_2}\right)hk+\dotsb$$ For this to be a circle,

1. coefficient of $hk$ must be $0$, and
2. coefficients of $h$ and $k$ should be equal.

Thus we have \begin{align*} \lambda_2a_1^2+\lambda_1a_2^2 & = \lambda_2b_1^2+\lambda_1b_2^2\\ a_1b_1\lambda_2+a_2b_2\lambda_1 & = 0 \end{align*}

Upon solving this, we get $$(a_1a_2)^2 = (b_1b_2)^2 \implies a_1a_2 = \pm b_1b_2.$$

Try to see why $a_1a_2=b_1b_2$ will not occur. Then the only thing left is $a_1a_2+b_1b_2=0$ which is the condition for perpendicularity of $l_1$ and $l_2$.

Answer 2

Ignoring the case of parallel (or coincident) lines, suppose $\ell_1$ and $\ell_2$ meet at the unique point $O$. Any circle about $O$ meets the lines at the vertices of a rectangle $\square ABCD$; it also meets the bisectors of the angles formed by those lines at the vertices of a square $\square WXYZ$ (because the bisectors are necessarily perpendicular).

Each vertex of $\square ABCD$ is at distance $0$ from either $\ell_1$ or $\ell_2$, and is at some common distance (say $k$) from the other of $\ell_1$ or $\ell_2$. Thus, "the sum of the squares of the distances to the lines" is a constant (namely, $k^2$) across all four points, which implies that $\bigcirc O$ must be one of the "locus-circles" determined by the lines. That sum must then also be constant across the vertices of $\square WXYZ$, as those vertices lie on that locus-circle; in particular, the sums for $W$ and $X$ alone should match. However, because $W$ is on an angle bisector, the sum of the squares of the distances from $W$ to both lines is just double the square of the distance to either line; likewise for $X$. We conclude that the distances from each of $W$ and $X$ to each of $\ell_1$ and $\ell_2$ all match, making one of the lines a parallel to segment $\overline{WX}$ and the other line the perpendicular bisector of that segment. $\square$