The equation of the circle is:
$x^2 + (y - 1)^2 = 1$,
which represents a circle with center $C(0, 1)$ and radius 1.
The midpoints of all chords passing through the origin lie on another circle. The center of this circle is the midpoint between the origin and the center of the given circle, and the radius is half the radius of the original circle.
$\left( \frac{0 + 0}{2}, \frac{0 + 1}{2} \right) = \left( 0, \frac{1}{2} \right)$.
Thus, the equation of the locus of midpoints is:
$x^2 + \left( y - \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^2$,
which simplifies to:
$x^2 + \left( y - \frac{1}{2} \right)^2 = \frac{1}{4}$.
Now, we need to find the points of intersection of the circle $x^2 + \left( y - \frac{1}{2} \right)^2 = \frac{1}{4}$ with the line $x + y = 1$.
First, solve for $y$ in terms of $x$ from the line equation:
$y = 1 - x$.
Substitute this into the equation of the circle:
$x^2 + \left( 1 - x - \frac{1}{2} \right)^2 = \frac{1}{4}$.
Simplifying:
$x^2 + \left( \frac{1}{2} - x \right)^2 = \frac{1}{4}$.
Expand the square:
$x^2 + \left( \frac{1}{4} - x + x^2 \right) = \frac{1}{4}$.
Simplifying:
$x^2 + \frac{1}{4} - x + x^2 = \frac{1}{4}$.
Combine like terms:
$2x^2 - x + \frac{1}{4} = \frac{1}{4}$.
Cancel the $\frac{1}{4}$ terms:
$2x^2 - x = 0$.
Factor the equation:
$x(2x - 1) = 0$.
Thus, $x = 0$ or $x = \frac{1}{2}$.
For $x = 0$, substitute into the line equation $x + y = 1$:
$0 + y = 1 \implies y = 1$.
Thus, one point of intersection is $P(0, 1)$.
For $x = \frac{1}{2}$, substitute into the line equation:
$\frac{1}{2} + y = 1 \implies y = \frac{1}{2}$.
Thus, the second point of intersection is $Q \left( \frac{1}{2}, \frac{1}{2} \right)$.
Now, we calculate the distance between $P(0, 1)$ and $Q \left( \frac{1}{2}, \frac{1}{2} \right)$ using the distance formula:
$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Substitute the coordinates of P and Q:
$PQ = \sqrt{\left( \frac{1}{2} - 0 \right)^2 + \left( \frac{1}{2} - 1 \right)^2} = \sqrt{\left( \frac{1}{2} \right)^2 + \left( - \frac{1}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Thus, the length of PQ is:
$\boxed{\frac{1}{\sqrt{2}}}$
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, -3)$ and eccentricity is $\frac{4}{5}$, is
In a two-dimensional coordinate system, it is proposed to determine the size and shape of a triangle ABC in addition to its location and orientation. For this, all the internal angles and sides of the triangle were observed. Further, the planar coordinates of point A and bearing/azimuth of line AB were known. The redundancy (\( r \)) for the above system will be equal to _________ (Answer in integer).
Match List-I with List-II: List-I