A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{8}\]
Correct Answer: C
Solution :
We have, the equation of circle \[{{(x-7)}^{2}}+{{(y+1)}^{2}}=25\] \[\Rightarrow \] \[{{x}^{2}}+49-14x+{{y}^{2}}+1+2y=25\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-14x+2y+25=0\] On comparing with\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c,\]we get \[2g=-14,2g=2,c=25\] \[\Rightarrow \] \[g=-7,f=1\] Now, \[{{g}^{2}}+{{f}^{2}}={{(-7)}^{2}}+{{1}^{2}}=50\] \[\Rightarrow \] \[2c=2\times 25=50\] \[\because \] \[{{g}^{2}}+{{f}^{2}}=2c\] Thus, the tangents drawn from the origin to the circle are perpendicular. Hence, angle between them is\[\frac{\pi }{2}\].
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