A) \[2\]
B) \[3\]
C) \[0\]
D) \[1\]
Correct Answer: B
Solution :
Given circles are \[{{x}^{2}}+{{y}^{2}}-y=0\] and \[{{x}^{2}}+{{y}^{2}}+y=0\] centres and radii of these circles are \[{{C}_{1}}\left( 0,\frac{1}{2} \right),\] \[{{C}_{2}}\left( 0,-\frac{1}{2} \right)\] and \[{{r}_{1}}=\frac{1}{2},\] \[{{r}_{2}}=\frac{1}{2}\] Now, \[{{C}_{1}}{{C}_{2}}=\sqrt{0+{{\left( \frac{1}{2}+\frac{1}{2} \right)}^{2}}}=1\] and \[{{r}_{1}}+{{r}_{2}}=\frac{1}{2}+\frac{1}{2}=1\] \[\because \] \[{{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}}\] It means that two circles touch each other externally. Hence, number of common tangents are 3.
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