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, \[{{g}_{1}}=-1,\]\[{{f}_{1}}=-1,\]\[{{c}_{1}}=-7\] \[{{g}_{2}}=2,\]\[{{f}_{2}}=1,\] \[{{c}_{2}}=k\]You need to login to perform this action.
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