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 \[{{y}_{2}}=4\]You need to login to perform this action.
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