A) 0
B) \[\pi /3\]
C) \[\pi /6\]
D) \[\pi /2\]
Correct Answer: D
Solution :
Let the equation of tangent\[~y=mx\] of the circle \[{{(x-1)}^{2}}+{{(y+1)}^{2}}=25\] \[\therefore \]Distance from centre of the circle to the tangent = radius of the circle \[\Rightarrow \] \[\frac{-1+m}{\sqrt{1+{{m}^{2}}}}=5\] \[\Rightarrow \] \[{{1}^{2}}+{{m}^{2}}-2m=25(1+{{m}^{2}})\] \[\Rightarrow \] \[24{{m}^{2}}+2m+24=0\] \[\Rightarrow \] \[12\,{{m}^{2}}+m+12=0\] \[\Rightarrow \] \[{{m}_{1}}{{m}_{2}}=-1\] Alternate Solution: It is clear from the figure that circle touches the coordinate axes, therefore tangent is perpendicular.You need to login to perform this action.
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