Circle Question 3
Question 3 - 2024 (01 Feb Shift 2)
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 $P Q$ is :
(1) $\frac{1}{\sqrt{2}}$
(2) $\sqrt{2}$
(3) $\frac{1}{2}$
(4) 1
Show Answer
Answer (1)
Solution
$\mathrm{m}{\mathrm{OM}} \cdot \mathrm{m}{\mathrm{CM}}=-1$
$\frac{\mathrm{k}}{\mathrm{h}} \cdot \frac{\mathrm{k}-1}{\mathrm{~h}}=-1$
$\therefore$ locus is $\mathrm{x}^{2}+\mathrm{y}(\mathrm{y}-1)=0$
$\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{y}=0$ $\mathrm{p}=\left|\frac{1 / 2}{\sqrt{2}}\right| \mathrm{p}=\frac{1}{2 \sqrt{2}}$
$\mathrm{PQ}=2 \sqrt{\mathrm{r}^{2}-\mathrm{p}^{2}}$
$=2 \sqrt{\frac{1}{4}-\frac{1}{8}}=\frac{1}{\sqrt{2}}$
