A) 1
B) 3
C) 2
D) None of these
Correct Answer: C
Solution :
\[{{x}^{2}}+{{y}^{2}}+2x+8y-23=0\]\[\therefore {{C}_{1}}(-1,-4),{{r}_{1}}=2\sqrt{10}\] Again \[{{x}^{2}}+{{y}^{2}}-4x-10y+9=0\] \ \[{{C}_{2}}(2,5),{{r}_{2}}=2\sqrt{5}\] Now \[{{C}_{1}}{{C}_{2}}\]=distance between centres. \[\therefore \] \[{{C}_{1}}{{C}_{2}}=\sqrt{9+81}=3\sqrt{10}=9.486\]and \[{{r}_{1}}+{{r}_{2}}=2(\sqrt{10}+\sqrt{5})=10.6\] \[{{r}_{1}}-{{r}_{2}}=2\sqrt{5}(\sqrt{2}-1)=2\times 2.2\times 0.4=4.4\times 0.4=1.76\] \[{{C}_{1}}{{C}_{2}}=2\sqrt{10}>{{r}_{1}}-{{r}_{2}}\] \[{{r}_{1}}-{{r}_{2}}<{{C}_{1}}{{C}_{2}}<{{r}_{1}}+{{r}_{2}}\Rightarrow \]Two tangents can be drawn.
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