Find the locus of the circumcenter of a triangle whose two sides are along the coordinate axes and the third side passes through the point of intersection of the line `a x+b y+c=0` and `l x+m y+n=0.`

Explanation

Let the equation of the third line be
`(ax+by+c)+lambda(lx+my+n)=0`
`"where" lambda " is a parameter. It meets the x-axis at " (-(lambdan+c)/(lambdal+a),0) " and the "`
`"y-axis at B" (0,-(lambdan+c)/(lambdam+b))`
The triangle OAB is a right-angled triangle. Its circumcenter is the midpoint of the hypotenuse. Let it be `(alpha, beta)`. Then,
`2alpha = -((lambdan+c)/(lambdal+a))`
`"and " 2beta = -((lambdan+c)/(lambdam+b))`
`Hence ,lambda = -(2alphaa+c)/(2alphal+n) =-(2betab+c)/(2betam+n)`
Hence, the locus of `(alpha, beta)` is
`(c+2ax)/(n+2lx) =(c+2by)/(n+2my)`
or 2xy(bl-ma)=x(an-lc)+y(mc-bn)