A) 37/2
B) 4/37
C) 21
D) 7
E) 39/4
Correct Answer: E
Solution :
The length of tangent from the point (1,2) to the circle\[{{x}^{2}}+{{y}^{2}}+x+y-4=0\]is \[\sqrt{1+4+1+2-4},\]ie,2. And the length of tangent from the point (1, 2) to the circle\[3{{x}^{2}}+3{{y}^{2}}-x-y-k=0\]is \[\sqrt{3+12-1-2-k},\]ie, \[\sqrt{12-k}\] \[\therefore \] \[\frac{2}{\sqrt{12-k}}=\frac{4}{3}\Rightarrow \frac{2\times 3}{4}=\sqrt{12-k}\] \[\Rightarrow \] \[\frac{9}{4}=12-k\] \[\Rightarrow \] \[k=12-\frac{9}{4}=\frac{48-9}{4}=\frac{39}{4}\]
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