Question

Two concentric circles of radii a and b (a >

b.are given. find the length of the chord of the larger circle which touches the smaller circle.

Answer 1

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Answer 2

Answer:

$\text{Length of chord=AB=}\frac{1}{2}(\sqrt{a^2-b^2})$

Step-by-step explanation:

Given two concentric circles of radii a and b(a>b).

we have to find the length of the chord of the larger circle which touches the smaller circle.

Let AB be the chord of the larger circle and tangent at C to smaller circle.

As, AB is tangent  and radius is perpendicular at tangent ∴ ∠OCB = 90°

Also, radius from the center perpendicularly bisect the chord.

⇒ $AC=CB=\frac{1}{2}AB$

In ΔOCB,

By Pythagoras theorem,

$OB^2=OC^2+CB^2$

⇒ $a^2=b^2+CB^2$

⇒ $CB=\sqrt{a^2-b^2}$

$\text{Length of chord=}AB=\frac{1}{2}(\sqrt{a^2-b^2})$