A) \[60{}^\circ \]
B) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{3}{5} \right)\]
D) \[135{}^\circ \]
Correct Answer: D
Solution :
Given circles are\[{{x}^{2}}+{{y}^{2}}-2x+8y+13=0\] and \[{{x}^{2}}+{{y}^{2}}-4x+6y+11=0.\] Here, \[{{C}_{1}}=(1,\,-4),\,{{C}_{2}}=(2,\,-3)\] \[\Rightarrow \] \[{{r}_{1}}=\sqrt{1+16-13}=2\] and \[{{r}_{2}}=\sqrt{4+9-11}=\sqrt{2}\] Now, \[d={{C}_{1}}{{C}_{2}}=\sqrt{{{(2-1)}^{2}}+{{(-3+4)}^{2}}}=\sqrt{2}\] \[\therefore \]\[\cos \theta =\frac{{{d}^{2}}-r_{1}^{2}-r_{2}^{2}}{2{{r}_{1}}{{r}_{2}}}=\frac{2-4-2}{2\times 2\times \sqrt{2}}=-\frac{1}{\sqrt{2}}\] \[\Rightarrow \] \[\theta =135{}^\circ \]
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