question_answer

The greatest distance of the point (10, 7) from the circle \[{{x}^{2}}+{{y}^{2}}-4x-2y-20=0\] is \[5\alpha ,\] then \[\alpha \] is equal to

A)  3                                            

B)  4                 

C)  5                                            

D)  6

Correct Answer: A

Solution :

 Centre of circle = (2, 1) and    radius \[{{V}_{2}}>{{V}_{4}}\] \[{{V}_{3}}\]       \[MgC{{l}_{2}}\to 2NaOH\to Mg{{(OH)}_{2}}\downarrow +2NaCl\] \[Mg{{(OH)}_{2}}\]         \[M{{g}^{2+}}\] = 10 \[O{{H}^{-}}\] At point (10, 7), \[{{K}_{sp}}\] \[Mg{{(OH)}_{2}}\] The point P is outside the given circle. Now, required largest distance = PA \[Mg{{(OH)}_{2}}\,\underset{s}{\mathop{M{{g}^{2+}}}}\,+\underset{2s}{\mathop{2O{{H}^{-}}}}\,\]      PA = AB + BP = 2 radius + (OP - radius) \[{{K}_{sp}}=4{{s}^{3}}\] \[s=\sqrt[3]{\frac{12\times {{10}^{-12}}}{4}}=1.44\times {{10}^{-4}}\] \[\therefore \]  \[[O{{H}^{-}}]=2s=2\times 1.44\times {{10}^{-4}}=2.88\times {{10}^{-4}}\]                            (given) \[1\,d{{m}^{3}}=1\,L=1000\,mL\]              \[\therefore \]