question_answer

The length of common chord of the circles \[{{x}^{2}}+{{y}^{2}}=12\]and \[{{x}^{2}}+{{y}^{2}}-4x+3y-2=0\], is                              [RPET 1990, 99]

A)            \[4\sqrt{2}\]                             

B)            \[5\sqrt{2}\]

C)            \[2\sqrt{2}\]                             

D)            \[6\sqrt{2}\]

Correct Answer: A

Solution :

 The equation of common chord \[\equiv {{S}_{1}}-{{S}_{2}}=0\] or \[4x-3y-10=0\] and centre of first circle is (0, 0). Therefore perpendicular from it on line is \[{{p}_{1}}=\frac{10}{5}=2\] and\[{{R}_{1}}=\sqrt{12}\]. Hence\[{{L}_{1}}{{L}_{2}}=2\sqrt{(R_{1}^{2}-p_{1}^{2})}=2\sqrt{(12-4)}=4\sqrt{2}\].