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, \[PQ=2PM=2\sqrt{{{({{C}_{1}}P)}^{2}}-{{({{C}_{1}}M)}^{2}}}\] \[=2\sqrt{9-{{\left( \frac{9}{\sqrt{13}} \right)}^{2}}}\] \[=2\sqrt{9-\frac{81}{13}}\] \[=2\sqrt{\frac{36}{13}}=\frac{12}{\sqrt{13}}\]You need to login to perform this action.
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