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\[{{L}_{1}}{{L}_{2}}=2\sqrt{(R_{1}^{2}-p_{1}^{2})}=2\sqrt{(12-4)}=4\sqrt{2}\].You need to login to perform this action.
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