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If two circles A, B of equal radii pass through the centres of each other, then what is the ratio of the length of the smaller arc to the circumference of the circle A cut off by the circle $\mathrm{B} ?$
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The correct answer is:
$\frac{1}{3}$
When two circles A and B of equal radii pass through the centres of each other, the angle made by arc of $\mathrm{B}$ at
the centre of $\mathrm{B}$ is $90^{\circ}$.
So, length of small $\operatorname{arc}$ of $\mathrm{B}=\frac{2 \pi \mathrm{r} 90^{\circ}}{360^{\circ}}=\frac{\pi \mathrm{r}}{2}$
Hence, circumference of A cut off by the circle $\mathrm{B}$
$=2 \pi r-\frac{\pi r}{2}=\frac{3 \pi r}{2}$
$\therefore \quad$ Required ratio $=\frac{\pi \mathrm{r} / 2}{3 \pi \mathrm{r} / 2}=\frac{1}{3}$
the centre of $\mathrm{B}$ is $90^{\circ}$.
So, length of small $\operatorname{arc}$ of $\mathrm{B}=\frac{2 \pi \mathrm{r} 90^{\circ}}{360^{\circ}}=\frac{\pi \mathrm{r}}{2}$
Hence, circumference of A cut off by the circle $\mathrm{B}$
$=2 \pi r-\frac{\pi r}{2}=\frac{3 \pi r}{2}$
$\therefore \quad$ Required ratio $=\frac{\pi \mathrm{r} / 2}{3 \pi \mathrm{r} / 2}=\frac{1}{3}$
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