Animated Solution for Mathematics - Circles: From the origin chords are drawn to the circle $(x-1{)}^2+y^2=1$. The equation of the locus of the mid-points of these chords is .........

The equation $(x-1{)}^2+y^2=1$ describes a circle with its center at $(1,0)$ and a radius of $1$.

By observing the equation, we see that the origin $(0,0)$ lies exactly on the boundary of this circle. This serves as the fixed starting point for all chords considered in this problem.

Expanding the circle equation $(x-1{)}^2+y^2=1$ yields:

This simplifies to the standard form:

Let $P(h,k)$ be the midpoint of a chord originating from $(0,0)$. We utilize the chord equation formula $T=S_1$, where $T$ represents the tangent-like expression at $(h,k)$ and $S_1$ is the value of the circle equation at $(h,k)$.

For the circle $x^2+y^2-2x=0$, we apply the standard replacement rules: $x^2\to xh$, $y^2\to yk$, and $x\to \frac{x+h}{2}$.

This gives us the expression for $T$:

Next, we calculate $S_1$ by substituting the midpoint $(h,k)$ into the circle equation:

Setting $T=S_1$, we obtain the equation of the chord:

Since every chord passes through the origin $(0,0)$, the coordinates $(0,0)$ must satisfy the chord equation. Substituting $x=0$ and $y=0$ into the equation above:

This simplifies to:

Rearranging the terms, we arrive at the final relation:

Replacing $(h,k)$ with the general coordinates $(x,y)$, we find the locus of the midpoints:

This result represents a circle with its center at $(\frac{1}{2},0)$ and a radius of $\frac{1}{2}$. The midpoints of all chords drawn from the origin to the original circle trace out this smaller, perfectly defined geometric path.