question_answer

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

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

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

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

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

Correct Answer: C

Solution :

The equations of two circles are                 \[{{S}_{1}}\equiv {{(x-a)}^{2}}+{{(y-b)}^{2}}={{c}^{2}}\] ... (i) and        \[{{S}_{2}}\equiv {{(x-b)}^{2}}+{{(y-a)}^{2}}={{c}^{2}}\] ... (ii) The equation of the common chord of these circle is                 \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \]\[{{(x-a)}^{2}}-{{(x-b)}^{2}}+{{(y-b)}^{2}}-{{(y-a)}^{2}}=0\] \[\Rightarrow \]\[(2x-a-b)(b-a)+(2y-b-a)(a-b)=0\] \[\Rightarrow \,2x-a-b-2y+b+a=0\] \[\Rightarrow x-y=0\] Let\[{{C}_{1}}\]and\[{{C}_{2}}\]be the centres of circles (i) and (ii) respectively. Then, the coordinates of\[{{C}_{1}}\] and\[{{C}_{2}}\]are\[(a,\,\,b)\]and\[(b,\,\,a)\]respectively.                 \[{{C}_{1}}M=\left| \frac{a-b}{\sqrt{1+1}} \right|=\frac{|a-b|}{\sqrt{2}}\] In right\[\Delta {{C}_{1}}PM\], we have,                 \[PM=\sqrt{{{C}_{1}}{{P}^{2}}-{{C}_{1}}{{M}^{2}}}=\sqrt{{{c}^{2}}-\frac{{{(a-b)}^{2}}}{2}}\] \[\therefore \]  \[PQ=2PM=2\sqrt{{{c}^{2}}-\frac{{{(a-b)}^{2}}}{2}}\]                 \[=\sqrt{4{{c}^{2}}-2{{(a-b)}^{2}}}\]