A) (6, 7)
B) \[(-6,7)\]
C) \[(6,-7)\]
D) \[(-6,-7)\]
Correct Answer: D
Solution :
The tangent at (1, 7) to. the parabola \[{{x}^{2}}=y-6\]is \[x(1)=\frac{1}{2}(y+7)-6\] [replacing\[{{x}^{2}}\to x{{x}_{1}}\]and \[2y\to y+{{y}_{1}}\]] \[\Rightarrow \] \[2x=y+7-12\] \[\Rightarrow \] \[y=2x+5\] ...(i) Which is also tangent to the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\] ie, \[{{x}^{2}}+{{(2x+5)}^{2}}+16x+12(2x+5)+c=0\] or \[5{{x}^{2}}+60x+85+c=0\] must have equal roots. \[\Rightarrow \]\[\alpha =\beta \]for above equation ie, \[\Rightarrow \]\[\alpha +\beta =-\frac{60}{5}\]or\[\alpha =-6\] (as\[\alpha =\beta \]) \[\therefore \] \[x=-6\]and \[y=2x+5=-7\] \[\Rightarrow \]Point of contact is\[(-6,-7)\].
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