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Two circles with radii $\sqrt{2} c m$ and 1 cm meet at a point A, whose centres lie on x-axis. The distance between their centres is 2 cm. The chord AC of the larger circle cuts the smaller circle at a point B and is bisected by that point. The length of chord AC is

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$\sqrt{\frac{7}{2}} c m$

$\frac{\sqrt{5}}{2} c m$

$\frac{\sqrt{3}}{2} c m$

$\sqrt{\frac{5}{2}} c m$

answer is B.

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Detailed Solution

Let origin be centre of larger circle and $O'$ be the centre of smaller circle.

Equation of (1) $x^{2} + y^{2} = 2$

Equation of (2) ${(x - 2)}^{2} + y^{2} = 1$

$∴ A (\frac{5}{4}, \frac{\sqrt{7}}{4})$

$x_{B} = \frac{x_{A} + x_{C}}{2}, y_{B} = \frac{y_{A} + y_{C}}{2}$

$∴ x_{C}^{2} + y_{C}^{2} = 2$

$∴ {(\frac{x_{C} + \frac{5}{4}}{2} - 2)}^{2} + {(\frac{y_{C} + \frac{\sqrt{7}}{4}}{2})}^{2} = 1$

$∴ A C = \sqrt{\frac{7}{2}}$

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