Animated Solution for Mathematics - Circles: If the circles $x^2+y^2+5Kx+2y+K=0$ and $2(x^2+y^2)+2Kx+3y-1=0$, $(K\in R)$, intersect at the points P and Q, then the line $4x+5y-K=0$ passes through P and Q for :

Imagine standing on a vast coordinate plane, watching two circles, $S_1$ and $S_2$, as they dance toward each other. They collide, creating two distinct points of intersection, $P$ and $Q$.

We are tasked with finding a value of $K$ such that the line $4x+5y-K=0$ acts as the bridge between these two points—the common chord.

The Hidden Geometry

In the world of coordinate geometry, the line passing through the intersection points of two circles is known as the radical axis. It is a line of perfect balance, where the power of a point with respect to both circles is equal.

To find it, we use the elegant subtraction $S_1-S_2=0$. However, this subtraction only works if the circles are normalized. If the coefficients of $x^2$ and $y^2$ are not identical, the subtraction leaves behind a quadratic mess instead of a clean, linear equation.

In our case, $S_1$ has coefficients of $1$, but $S_2$ has coefficients of $2$. We must normalize $S_2$ by dividing by $2$, transforming it into:

With both circles normalized, we subtract them:

Watch as the quadratic terms vanish, leaving us with the linear equation of the common chord:

This is the heart of the problem. We are told this line is identical to $4x+5y-K=0$. For two lines to be identical, they must be proportional. This means the ratio of their coefficients must be equal:

The Final Hurdle

This is where many students stumble. We solve the first part of the proportion:

It is tempting to stop here, to declare victory and move on. But a true JEE aspirant knows that the third ratio is the ultimate judge. We must check if $K=0.1$ satisfies the final part of the proportion:

Substituting $K=0.1$, we get:

Since $0.1 eq -6$, our system is inconsistent. The lines can never be the same.

Thus, we conclude with mathematical certainty: there is no value of $K$ that satisfies the condition. This problem teaches us that in mathematics, as in life, consistency is everything. Do not just find a solution; verify that it holds true across all conditions.