Question

If q is the angle of intersection of two circle x² + y² = a² and (x – c)² + y² = b², then the length of common chord of two circle is –

• A. $\frac{ab}{\sqrt{a^{2} + b^{2} - 2ab\cos\theta}}$
• B. $\frac{2ab}{\sqrt{a^{2} + b^{2} - 2ab\cos\theta}}$
• C. $\frac{2ab\sin\theta}{\sqrt{a^{2} + b^{2} - 2ab\cos\theta}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{2ab\sin\theta}{\sqrt{a^{2} + b^{2} - 2ab\cos\theta}}$

Answer 2

We have the two circles x² + y² = a², (x – c)² + y² = b² radius of first circle = a

Let ŠOPM = a, ŠAPM = b, \ ŠOPA = a + b = q

Let PQ = x

radius of second circle = b

cos a = PM/a = x/2a, cos b = x/2b

Now cos q = cos (a + b) = cos a cos b – sin a sin b

\ sin a sin b = cos a cos b – cos q

Squaring of both sides, we get

sin² a sin² b = cos² a cos² b + cos² q – 2 cos q cos a cos b

1 – cos² a – cos² b + cos² a cos² b = cos² a cos² b + cos² q –

2 cos q cos a cos b

Ž \ sin²q = cos² a + cos² b – 2 cos q cos a cos b

sin² q = $\frac{x^{2}}{4a^{2}}$+$\frac{x^{2}}{4b^{2}}$–$\frac{2.x^{2}}{4ab}$. cos q

Ž 4a²b² sin² q = x² (b² + a² – 2ab cos q)

\ x = $\frac{2ab\sin\theta}{\sqrt{a^{2} + b^{2} - 2ab\cos\theta}}$.