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Question: Length of the common chord of the circles (x – 1)<sup>2</sup> + (y + 1)<sup>2</sup> = c<sup>2</sup>...

Length of the common chord of the circles

(x – 1)2 + (y + 1)2 = c2 and (x + 1)2 + (y – 1)2 = c2 is –

A

12c2−2\frac{1}{2}\sqrt{c^{2} - 2}

B

c2−2\sqrt{c^{2} - 2}

C

2c2−22\sqrt{c^{2} - 2}

D

c +2

Answer

2c2−22\sqrt{c^{2} - 2}

Explanation

Solution

Equation of the common chord AB of the given circles is

(x – 1)2 + (y + 1)2 – (x + 1)2 – (y – 1)2 = 0 Ž y = x

Let C1 (1, –1) be the centre of the first circle and M be the mid-point of AB, then C1A= c, C1 M = ∣1+12∣\left| \frac{1 + 1}{\sqrt{2}} \right| = 2\sqrt{2}

and AB = 2AM = 2(C1A)2–(C1M)22\sqrt{(C_{1}A)^{2}–(C_{1}M)^{2}}

= 2c2−22\sqrt{c^{2} - 2}.