Let the locus of the midpoints of the chords | vedclass

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(A) Let $M(h, k)$ be the midpoint of a chord drawn from the origin $O(0, 0)$.The center of the circle $x^2+(y-1)^2=1$ is $C(0, 1)$.Since $CM \perp OM$

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

Vedclass · Answer

(A) Let $M(h, k)$ be the midpoint of a chord drawn from the origin $O(0, 0)$.The center of the circle $x^2+(y-1)^2=1$ is $C(0, 1)$.Since $CM \perp OM$,the product of their slopes is $-1$:$\left(\frac{k-1}{h-0}\right) \cdot \left(\frac{k-0}{h-0}\right) = -1$$\frac{k(k-1)}{h^2} = -1$$k^2-k = -h^2$$h^2+k^2-k = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $x^2+y^2-y=0$.This is a circle with center $(0, 1/2)$ and radius $r = 1/2$.The line is $x+y-1=0$.The perpendicular distance $p$ from the center $(0, 1/2)$ to the line $x+y-1=0$ is:$p = \frac{|0 + 1/2 - 1|}{\sqrt{1^2+1^2}} = \frac{1/2}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.The length of the chord $PQ$ is $2\sqrt{r^2-p^2}$:$PQ = 2\sqrt{(1/2)^2 - (1/(2\sqrt{2}))^2} = 2\sqrt{1/4 - 1/8} = 2\sqrt{1/8} = 2 \cdot \frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.

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:

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