Let the locus of the midpoints of the chords of the circle drawn from the origin intersect the line x + y = 1 at P and Q. Then, the length of PQ is : [2024]

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Solution

Q.
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 : [2024]

1 $\frac{1}{2}$

2 1

3 $\frac{1}{\sqrt{2}}$

4 $\sqrt{2}$

Ans.
(3)

Let $(x_{1}, y_{1})$ be the mid point of chords.

So, equation of chord of the circle $x^{2} + y^{2} - 2 y = 0$ is,

$x \cdot x_{1} + y \cdot y_{1} - (y + y_{1}) = x_{1}^{2} + y_{1}^{2} - 2 y_{1}$

The chord is passing through origin

$∴ - y_{1} = x_{1}^{2} + y_{1}^{2} - 2 y_{1}$

$x_{1}^{2} + y_{1}^{2} - y_{1} = 0$ ... (i)

Now, (i) intersects the line x + y = 1

$∴ {(1 - y_{1})}^{2} + y_{1}^{2} - y_{1} = 0$

$\Rightarrow 1 + y_{1}^{2} - 2 y_{1} + y_{1}^{2} - y_{1} = 0 \Rightarrow 2 y_{1}^{2} - 3 y_{1} + 1 = 0$

$\Rightarrow (2 y_{1} - 1) (y_{1} - 1) = 0 \Rightarrow y_{1} = \frac{1}{2} or y_{1} = 1$

If $y_{1} = \frac{1}{2}$ , then $x_{1} = \frac{1}{2}$ [From (i)]

If y = 1, then $x_{1} = 0$

So, $P = (\frac{1}{2}, \frac{1}{2})$ and Q = (1, 0) $∴ P Q = \sqrt{{(\frac{1}{2})}^{2} + {(\frac{- 1}{2})}^{2}} = \frac{1}{\sqrt{2}}$

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