Question

Let the locus of the midpoints of the chords of the 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:

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

The Correct Option is A

Given:  

The point \( P(3, 4, 9) \) lies on the line with the parametric equations: \[ \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1} = \lambda \]

Find: Any point on the line, represented as \( Q \left( 3\lambda + 1, 2\lambda - 1, \lambda + 2 \right) \).

Step 1: Vector form of line \( \overrightarrow{PQ} \):

From \( P(3, 4, 9) \) to \( Q(3\lambda + 1, 2\lambda - 1, \lambda + 2) \), the direction vector is: \[ \overrightarrow{PQ} = \langle 3\lambda - 2, 2\lambda - 5, \lambda - 7 \rangle \]

Step 2: Equating components of the direction vector:

The equation becomes: \[ \langle 3\lambda - 2, 2\lambda - 5, \lambda - 7 \rangle \times \langle 3, 2, 1 \rangle = 0 \]

Expanding the determinant: \[ 9\lambda - 6 + 4\lambda - 10 + \lambda - 7 = 0 \] Simplifying: \[ 14\lambda - 23 = 0 \]

Step 3: Solving for \( \lambda \):

\[ \lambda = \frac{23}{14} \]

Step 4: Finding the coordinates of point \( Q \):

The coordinates of \( Q \) are: \[ Q \left( \frac{83}{14}, \frac{32}{14}, \frac{51}{14} \right) \]

Step 5: Parametric coordinates of \( P \) and \( Q \):

For point \( P(3, 4, 9) \), the parametric coordinates are: \[ 3 + \alpha_1 = \frac{83}{14} \quad \Rightarrow \quad x_1 = \frac{62}{7} \] \[ 4 + \beta_1 = \frac{32}{14} \quad \Rightarrow \quad y_1 = \frac{4}{7} \] \[ 9 + \gamma_1 = \frac{51}{14} \quad \Rightarrow \quad z_1 = \frac{12}{7} \]

Step 6: Final Result:

Now, we calculate: \[ 14 (\alpha + \beta + \gamma) = 14 \times \left( \frac{62}{7} + \frac{4}{7} - \frac{12}{7} \right) = 108 \]

Conclusion: The calculation is consistent, and the value is verified.

Answer 2

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:

$\boxed{\frac{1}{\sqrt{2}}}$