A) \[{{c}^{2}}+6c+7=0\]
B) \[{{c}^{2}}-7c+6=0\]
C) \[{{c}^{2}}+7c+6=0\]
D) \[{{c}^{2}}-6c+7=0\]
Correct Answer: A
Solution :
Tangent at \[\left( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)\] on circle \[{{x}^{2}}+{{y}^{2}}=1\] is \[x+y=\sqrt{2}\] \[\therefore \] Slope of tangent m = 1 for circle \[{{\left( x-3 \right)}^{2}}+{{y}^{2}}=1\] \[\because \] Any tangent of circle \[{{(x-3)}^{2}}+{{y}^{2}}=1\] is \[y=mx-3m\pm \sqrt{1+{{m}^{2}}}\] \[\therefore \,\,\,\,\,\,c+3m=\pm \sqrt{1+{{m}^{2}}}\] \[\therefore \,\,m=1\] \[\Rightarrow \] \[{{c}^{2}}+6c+7=0\]
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