Find locus of point of intersection of $x+2y+\lambda(x-2y)=0$, $x+y-2+\beta(x-2)=0$, if these are always perpendicular to each other.
My attempt is as follows:-
$x+2y+\lambda(x-2y)=0$, if we see carefully this represents the family of lines which pass through the intersection of $x+2y=0$ and $x-2y=0$ which is $(0,0)$
In the same way $x+y-2+\beta(x-2)=0$ represents family of lines which pass through the intersection of $x+y-2=0$ and $x-2=0$ which is $(2,0)$
Now we have to find the locus of point of intersection of a line from first family and a line from second family which are perpendicular to each other.
$$\angle ABC=90^\circ$$
Now one solution which comes to mind is that $(h,k)$ lies on circle and $A,B$ are the endpoints of diameter. In that case locus of $(h,k)$ will be
$$x(x-2)+y^2=0$$
Now I am wondering if there is any other locus which also satisfies the given properties?
No, the circle with diametric ends as such is the only region where the lines can intersect at $90^o$. Hence the locus you arrived with is the only correct answer.