A) 0
B) 1
C) 2
D) 3
Correct Answer: A
Solution :
We have the equation of circles \[{{x}^{2}}+{{(y-1)}^{2}}=9\] ...(i) and \[{{(x-1)}^{2}}+{{y}^{2}}=25\] ...(ii) For the circle (i), Centre\[{{C}_{1}}(0,1)\]and radius \[{{r}_{1}}=3\] For the circle (ii), Centre\[{{C}_{2}}(1,0)\]and radius\[{{r}_{2}}=5\] Now, \[{{C}_{1}}{{C}_{2}}=\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\] and \[{{r}_{2}}-{{r}_{1}}=2\] Clearly, \[{{C}_{1}}{{C}_{2}}<{{r}_{2}}-{{r}_{1}}\] Therefore, one circle lies entirely inside the other. Hence, there is no common tangent to the given circles.
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