A Triangle is formed by the lines $X + Y = 0, X - Y = 0$ and $LX + MY = 1$ where $L^2 + M^2= 1$. Find the locus of circumcenter of triangle so formed.

Question

Using basic Geometry I have gotten the coordinates of the circumcenter in terms of $L$ and $M$, I don't have any idea how to proceed further.

To obtain the coordinates, solve the three equations taking $2$ at a time. Since the triangle is a right triangle the circumcenter is on the midpoint of the hypotenuse. After solving I got, $C= (L/L^2 - M^2 ,M/L^2 - M^2)$. How do I find a locus for this point.

Thanks in advance.

Answer 1

Let's call $x=\frac{L}{L^2-M^2}$ and $y=\frac{M}{L^2-M^2}$. Then $$x^2-y^2=\frac1{L^2-M^2}$$and $$x^2+y^2=\frac{M^2+L^2}{(L^2-M^2)^2}=\frac{1}{(L^2-M^2)^2}=(x^2-y^2)^2$$

From here $$x^4-2x^2y^2+y^4-x^2-y^2=0$$This looks something like

Answer 2

The intersection between the sides $x+y=0$, $x-y=0$ and the third side $lx+my=1$ are $(\frac1{l-m}, -\frac1{l-m})$ and $(\frac1{l+m}, \frac1{l+m})$, respectively. The coordinates of the circumcenter is the intersection between

$$y+\frac1{2(l-m)}=x- \frac1{2(l-m)}$$ $$y-\frac1{2(l+m)}=-x+ \frac1{2(l+m)}$$

which are the lines perpendicular to the two sides and passing their midpoints.

Rearrange the two equations,

$$x+y=\frac1{l+m},\>\>\>\>\>x-y=\frac1{l-m}$$

and express

$$l=\frac{x}{x^2-y^2},\>\>\>\>\>m=-\frac{y}{x^2-y^2}$$

Apply the given $l^2+m^2=1$ to obtain the locus equation of the circumcenter

$$x^2+y^2=(x^2-y^2)^2$$