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Home  >>  EAMCET  >>  Mathematics
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The points (3,-4) lies on both the cirlces $x^2+y^2-2x+8y+13=0$ and $x^2+y^2-4x+6y+11=0.$ Then the angle between the circles is :

\[\begin {array} {1 1} (1)\;60^{\circ} & \quad (2)\;\tan ^{-1} \bigg(\frac{1}{2}\bigg) \\ (3)\;\tan ^{-1} \bigg(\frac{3}{5}\bigg) & \quad (4)\;135^{\circ} \end {array}\]
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$(4)\;135^{\circ} $
answered Nov 7, 2013 by pady_1
 
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