A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
[d] Equations of the circles are \[{{x}^{2}}+{{y}^{2}}=4\] ?. (1) and \[{{x}^{2}}+{{y}^{2}}=2x+2y\] ?. (2) Centre of (1) is \[{{C}_{1}}\equiv (0,\,\,0)\]; Radius of (1) \[={{r}_{1}}=2;\] Centre of (2) is \[{{C}_{2}}\equiv (1,\,\,1);\] Radius of (2) \[={{r}_{2}}=\sqrt{2}\]\[d=\] distance between centres \[={{C}_{1}}{{C}_{2}}=\sqrt{1+1}=\sqrt{2}\] If \[\theta \] is the angle of intersection of two circles, then \[\cos \theta =\frac{{{r}^{2}}_{1}+{{r}^{2}}_{2}-{{d}^{2}}}{2{{r}_{1}}{{r}_{2}}}=\frac{{{(2)}^{2}}+{{(\sqrt{2})}^{2}}-{{(\sqrt{2})}^{2}}}{2.2.\sqrt{2}}=\frac{1}{\sqrt{2}}\] \[\therefore \,\,\,\,\theta =\frac{\pi }{4}\]You need to login to perform this action.
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