If the base of a triangle and the ratio of the lengths of the other two unequal sides are given, then the vertex lies on $a/an$

Let the base of the triangle be the line segment joining$B (0, 0)$ and $C (a, 0)$ and let the vertex be $A (h, k),$ where a is fixed. Also, let

$\begin{array}{l}\frac{AB}{AC}=\lambda ,\lambda \ne 1\\ \Rightarrow A{B}^2={\lambda }^{2}A{C}^2\\ \Rightarrow h^2+k^2={\lambda }^2\left\{(h-a{)}^2+k^2\right\}\\ \Rightarrow h^2\left(1-{\lambda }^2\right)+k^2\left(1-{\lambda }^2\right)+2a{\lambda }^{2}h-a^2{\lambda }^2=0\end{array}$

So,$A (h, k)$ lies on

$x^2+y^2+2a\frac{{\lambda }^2}{1-{\lambda }^2}x-\frac{a^2{\lambda }^2}{1-{\lambda }^2}=0$ which is a circle