question_answer

A square is inscribed in the circle \[{{x}^{2}}+{{y}^{2}}-6x+8y-103=0\]with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

A) 13                    

B)                                       \[\sqrt{41}\]

C)               6                                             

D)               \[\sqrt{137}\]

Correct Answer: B

Solution :

Given, \[{{x}^{2}}+{{y}^{2}}-6x+8y-103=0\]Centre is (3,-4) and radius is \[\sqrt{9+16+103}=8\sqrt{2}\]Let the coordinates of A, B, C and D are respectively \[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})\]and\[({{x}_{1}},{{y}_{2}}).\] \[\therefore \]\[\frac{{{x}_{1}}+{{x}_{2}}}{2}=3\]and\[\frac{{{y}_{1}}+{{y}_{2}}}{2}=-4\] \[\Rightarrow \]\[{{x}_{1}}+{{x}_{2}}=6\]                                             ?(i) and\[{{y}_{1}}+{{y}_{2}}=-8\]                                               ?(ii) Also,\[{{x}_{2}}-{{x}_{1}}={{y}_{2}}-{{y}_{1}}\]                             ?(iii)\[[\because AB=BC]\] \[AC=16\sqrt{2}\] \[\Rightarrow \]\[\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}=16\sqrt{2}\] \[\Rightarrow \]\[\sqrt{2{{({{x}_{2}}-{{x}_{1}})}^{2}}}=16\sqrt{2}\]                 [Using (iii)] \[\Rightarrow \]\[{{x}_{2}}-{{x}_{1}}=16\]                                             ...(iv) Solving (i) and (iv), we get \[{{x}_{1}}=-5\]and \[{{x}_{2}}=11\] Similarly, we can get \[{{y}_{1}}=-12\]and \[{{y}_{2}}=4\]