A) one
B) four
C) two
D) three
Correct Answer: C
Solution :
Given circles are \[{{x}^{2}}+{{y}^{2}}-8x-2y+1=0\] and\[{{x}^{2}}+{{y}^{2}}+6x+8y=0\] Their centres and radius are \[{{C}_{1}}(4,1){{r}_{1}}=\sqrt{16}=4\] \[{{C}_{2}}(-3,-4),{{r}_{2}}=\sqrt{25}=5\] \[{{r}_{1}}-{{r}_{2}}=-1,{{r}_{1}}+{{r}_{2}}=9\] Since, \[{{r}_{1}}-{{r}_{2}}<{{C}_{1}}{{C}_{2}}<{{r}_{1}}+{{r}_{2}}\] \[\therefore \]Number of common tangents = 2
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