Animated Solution for Mathematics - Conic Sections: If two distinct chords, drawn from the point $(p,q)$ on the circle $x^2+y^2=px+qy$ (where $pq\ne 0$) are bisected by the x-axis, then

We are given a circle with the equation:

The point $P(p,q)$ lies on this circle. We are tasked with finding the condition such that two distinct chords drawn from $P$ are bisected by the $x$-axis.

Let the midpoint of such a chord be $M(h,0)$, since the midpoint must lie on the $x$-axis.

To find the equation of a chord with a given midpoint $(h,k)$, we use the standard formula $T=S_1$. For the given circle, the equation of the chord with midpoint $(h,0)$ is:

Simplifying this expression, we obtain:

This formula is a lifesaver because it bypasses the need to find the slope of the chord or the coordinates of the endpoints. It directly provides the equation of the chord in terms of its midpoint.

Since the chord must pass through the point $P(p,q)$, we substitute $x=p$ and $y=q$ into the chord equation:

Multiplying the entire equation by $2$ to clear the denominators, we get:

Rearranging the terms into a standard quadratic equation in $h$, we arrive at:

For two distinct chords to exist, there must be two distinct midpoints. This implies that the quadratic equation in $h$ must have two distinct real roots.

For a quadratic equation $a{h}^2+bh+c=0$ to have two distinct real roots, the discriminant $D=b^2-4ac$ must be strictly greater than zero. Here, $a=2$, $b=-3p$, and $c=p^2+q^2$.

Calculating the discriminant:

Setting $D>0$ yields the final condition: $p^2>8q^2$

This is the beautiful, elegant condition we were looking for. It represents the culmination of our logical journey through the geometry of the circle.