Answer for question no.24

Let O be the centre of the concentric circles. AB is the chord of the larger circle and tangent to the smaller circle at C.

Given, OC = b and OB = a.

AB is the tangent to the smaller circle.

∴ ∠OCB = 90° (Radius is perpendicular to the tangent at point of contact)

OD ⊥ AB,

$∴ AC = BC = \frac{AB}{2} (Perpendicular from the centre to the chord bisects the chrod)$

In ΔOCB,

OB²= OC² + BC²

∴ BC² = OB² – OC² = a² – b²

$\Rightarrow BC = \sqrt{a^{2} - b^{2}} ∴ AB = 2BC = 2 \sqrt{a^{2} - b^{2}}$

Thus, the length of chord AB is $2 \sqrt{a^{2} - b^{2}}$ .