question_answer

The length of the common chord of the circles \[{{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}}\]and \[{{(x-b)}^{2}}+{{(y-a)}^{2}}={{c}^{2}}\], is

A)            \[\sqrt{4{{c}^{2}}-2{{(a-b)}^{2}}}\]                                 

B)            \[\sqrt{4{{c}^{2}}+2{{(a-b)}^{2}}}\]

C)            \[\sqrt{4{{c}^{2}}-2{{(a+b)}^{2}}}\]                               

D)            \[\sqrt{4{{c}^{2}}+2{{(a+b)}^{2}}}\]

Correct Answer: A

Solution :

           \[{{C}_{1}}(a,\ b),\ {{C}_{2}}(b,\ a),\ {{r}_{1}}={{r}_{2}}=c\]                    \[\therefore \]\[{{C}_{1}}P=\frac{1}{2}\sqrt{{{a}^{2}}+{{b}^{2}}+{{a}^{2}}+{{b}^{2}}-4ab}\]                    Length of common chord                    \[=2\text{ }{{\left[ {{c}^{2}}-\frac{1}{4}\left\{ 2({{a}^{2}}+{{b}^{2}})-4ab \right\} \right]}^{1/2}}\]                              \[=2\text{ }{{\left( \frac{2{{c}^{2}}-{{a}^{2}}-{{b}^{2}}+2ab}{2} \right)}^{1/2}}=\sqrt{4{{c}^{2}}-2{{(a-b)}^{2}}}\].