question_answer

If the circles \[{{x}^{2}}+{{y}^{2}}-2x-2y-7=0\]and \[{{x}^{2}}+{{y}^{2}}+4x+2y+k=0\] cut orthogonally, then the length of the common chord of the  circles is                                               

A)  \[\frac{12}{\sqrt{13}}\]                                

B)  \[2\]

C)  \[5\]                                    

D)  \[8\]

Correct Answer: A

Solution :

Given, circles are \[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-2x-2y-7=0\] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}+4x+2y+k=0\] Here,     \[{{g}_{1}}=-1,\]\[{{f}_{1}}=-1,\]\[{{c}_{1}}=-7\]                 \[{{g}_{2}}=2,\]\[{{f}_{2}}=1,\] \[{{c}_{2}}=k\]                 Equation of common chord is \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-2x-2y-7-{{x}^{2}}-{{y}^{2}}-4x\]                                                 \[-2y-k=0\] \[\Rightarrow \]               \[-6x-4y-7-k=0\] \[\Rightarrow \]               \[6x+4y+7+k=0\]              ?..(i) Since, \[{{S}_{1}}\] and \[{{S}_{2}}\] cut orthogonally \[\therefore \]  \[2({{g}_{1}}{{g}_{2}}+{{f}_{1}}{{f}_{2}})={{c}_{1}}+{{c}_{2}}\] \[\Rightarrow \]               \[2(-2-1)=-7+k\] \[\Rightarrow \]                               \[-6+7=k\] \[\Rightarrow \]                               \[k=1\] Then, from Eqs. (i), we get                 \[6x+4y+8=0\] Now, length of the common chord                 \[{{r}_{1}}=\sqrt{1+1+7}=3\]                 \[{{c}_{1}}=(1,1)\] Let \[{{C}_{1}}M\] = perpendicular distance from centre \[{{C}_{1}}(1,1)\] to the common chord                 \[6x+4y+8=0\] \[\Rightarrow \]  \[{{C}_{1}}M=\frac{|6+4+8|}{\sqrt{{{6}^{2}}+{{4}^{2}}}}=\frac{|18|}{\sqrt{{{5}^{2}}}}=\frac{18}{2\sqrt{13}}=\frac{9}{\sqrt{13}}\] Now, \[PQ=2PM=2\sqrt{{{({{C}_{1}}P)}^{2}}-{{({{C}_{1}}M)}^{2}}}\]                                 \[=2\sqrt{9-{{\left( \frac{9}{\sqrt{13}} \right)}^{2}}}\]                                 \[=2\sqrt{9-\frac{81}{13}}\]                                 \[=2\sqrt{\frac{36}{13}}=\frac{12}{\sqrt{13}}\]