Conic Sections Question 229
Question: The angle of intersection of the circles $ x^{2}+y^{2}=4 $ and $ x^{2}+y^{2}=2x+2y $ is
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{6} $
D) $ \frac{\pi }{4} $
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Answer:
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_1C_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}}{2r_1r_2}=\frac{{{(2)}^{2}}+{{(\sqrt{2})}^{2}}-{{(\sqrt{2})}^{2}}}{2.2.\sqrt{2}}=\frac{1}{\sqrt{2}} $
$ \therefore \theta =\frac{\pi }{4} $
BETA
