Question

Two concentric circles are of radii a and b (a>b). Length of the chord of the larger circle which touches the small circle is

(A) \(2\sqrt{a^2-b^2}\)

(B) \(\sqrt{a^2-b^2}\)

(C) \(2\sqrt{a^2+b^2}\)

(D) \(\sqrt{a^2+b^2}\)

Answer 1

Correct option is: (A)  2 \(\sqrt {a^2-b^2}\).

Let PQ is chord of larger circle which touches the smaller circle at point R.

\(\therefore\) OR = Radius of smaller circle = b.

OP = Radius of larger circle = a

\(\because\) PQ is tangent of smaller circle and OR is radius drawn through the point of contact R.

\(\therefore\) OR is perpendicular to PQ.

Also OR is perpendicular drawn from the centre to chord PQ of larger circle.

\(\therefore\) OR bisects PQ

i.e. PR = QR = \(\frac {PQ}{2}\)

= PQ = 2PR ....(1)

Now, in right \(\triangle\) PRO

\(OP^2 = OP^2 + PR^2\)

= \(PR^2 = OP^2 = a^2 -b^2\)

= PR = \(\sqrt {a^2-b^2}\)

\(\therefore\) PQ = 2 \(\sqrt {a^2-b^2}\) (From (1))

Hence, the length of chord of larger circle is 2 \(\sqrt {a^2-b^2}\).

Answer 2

Correct option is: (A) \(2\sqrt{a^2-b^2}\)