Animated Solution for Mathematics - Circles: If a circle passes through the points of intersection of the coordinate axes with the lines $\lambda x-y+1=0$ and $x-2y+3=0$, then the value of $\lambda =.........$

We are given two lines, $L_1:\lambda x-y+1=0$ and $L_2:x-2y+3=0$. Our objective is to determine the value of the parameter $\lambda$ such that the four points where these lines intersect the coordinate axes are concyclic.

First, we identify the intersection points for $L_1$: Setting $y=0$, we find the x-intercept: $x=-\frac{1}{\lambda }$. Thus, $A=(-\frac{1}{\lambda },0)$. Setting $x=0$, we find the y-intercept: $y=1$. Thus, $C=(0,1)$.

Next, we identify the intersection points for $L_2$: Setting $y=0$, we find the x-intercept: $x=-3$. Thus, $B=(-3,0)$. Setting $x=0$, we find the y-intercept: $y=\frac{3}{2}$. Thus, $D=(0,\frac{3}{2})$.

For the four points $A,B,C,$ and $D$ to lie on a single circle, we utilize the Power of a Point theorem relative to the origin $O(0,0)$. Since the coordinate axes act as intersecting chords of the circle meeting at the origin, the product of the segments on the x-axis must equal the product of the segments on the y-axis.

This condition is expressed as:

Substituting the magnitudes of the intercepts into this equation, we obtain:

Simplifying the product of the intercepts, we have:

Dividing both sides by $3$, we arrive at:

Solving for the parameter, we find the final result:

By leveraging the geometric properties of the axes, we avoided complex algebraic systems and arrived at the solution efficiently. This method remains a vital tool for solving concyclic problems in coordinate geometry.