A) 4
B) 2
C) 3
D) 1
Correct Answer: C
Solution :
The given equation of circles \[{{x}^{2}}+{{y}^{2}}-6x-8y+9=0\] and \[{{x}^{2}}+{{y}^{2}}=1\] Radius and centre of both circles is \[{{C}_{1}}\to (3,4),{{R}_{1}}=\sqrt{9+16-9}=4\] \[{{C}_{2}}\to (0,0),\,\,{{R}_{2}}=1\] \[{{C}_{1}}{{C}_{2}}=\sqrt{{{(3-0)}^{2}}+{{(4-0)}^{2}}}=\sqrt{9+16}=5\] \[{{R}_{1}}+{{R}_{2}}=4+1=5\] \[\because \] \[{{C}_{1}}{{C}_{2}}={{R}_{1}}+{{R}_{2}}\] In this case, two direct tangent are real and distinct while the transverse tangents are coincident. So, number of comman tangents = 3
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