Animated Solution for Mathematics - Circles: Let the locus of the mid points of the chords of circle $x^2+(y-1{)}^2=1$ drawn from the origin intersect the line $x+y=1$ at $P$ and $Q$. Then, the length of $PQ$ is :

Imagine you are standing at the origin of a coordinate plane, looking at a circle defined by the equation $x^2+(y-1{)}^2=1$. This circle is perfectly centered at $(0,1)$ with a radius of $1$.

Notice how the origin itself sits right on the edge of this circle. We are going to draw infinite chords from the origin to various points on this circle.

Each chord has a midpoint, and as we draw more and more chords, these midpoints begin to dance across the plane, tracing out a hidden path. Our goal is to find the equation of this path, known as the locus, and then see where it intersects the line $x+y=1$.

To find this locus, we need a geometric anchor. Let $M(h,k)$ be the midpoint of any chord drawn from the origin $O(0,0)$.

There is a beautiful, ironclad rule in geometry: the line segment connecting the center of a circle to the midpoint of a chord is always perpendicular to that chord. So, if $C$ is the center $(0,1)$ and $M$ is our midpoint $(h,k)$, then the line $CM$ is perpendicular to the line $OM$.

The slope of the chord $OM$ is:

The slope of the radius $CM$ is:

Since these lines are perpendicular, their slopes must satisfy the condition $m_1\cdot m_2=-1$.

Now, let us perform the calculation. Multiplying the slopes gives us:

This simplifies to $k(k-1)=-h^2$, which rearranges into $h^2+k^2-k=0$. By replacing the specific coordinates $(h,k)$ with the general variables $(x,y)$, we reveal the equation of our locus:

This is not just any equation; it is the equation of another circle! This new circle has its center at $(0,\frac{1}{2})$ and a radius of $\frac{1}{2}$.

Now that we have our locus, we need to find where it meets the line $x+y=1$. We can rewrite the line equation as $x=1-y$ and substitute it into our locus equation:

Expanding this, we get $(1-2y+y^2)+y^2-y=0$, which simplifies to:

Factoring this quadratic, we find $(2y-1)(y-1)=0$. This gives us two solutions: $y=1$ and $y=\frac{1}{2}$.

Correspondingly, when $y=1$, $x=0$, giving us point $P(0,1)$. When $y=\frac{1}{2}$, $x=\frac{1}{2}$, giving us point $Q(\frac{1}{2},\frac{1}{2})$.

We have our two points: $P(0,1)$ and $Q(\frac{1}{2},\frac{1}{2})$. The final step is to calculate the distance between them using the distance formula:

This becomes:

The length of the segment $PQ$ is exactly $\frac{1}{\sqrt{2}}$.