question_answer

If the tangent at (1, 7) to the curve \[{{x}^{2}}=y-6\] touches   the   circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\] then the value of c is: [JEE Main Online 08-04-2018]

A)  85                              

B)  95

C)  195                            

D)  185

Correct Answer: B

Solution :

                                  \[{{x}^{2}}=y-6\]                 tangent at\[\text{P(1, 7)}\]                 \[\text{x}\text{.1=}\left( \frac{y+7}{2} \right)-6\]                 \[\Rightarrow \text{2x -- y + 5 = 0 (eq}\text{. of tangent)}\])                 \[r=\sqrt{64+36-c}\]                 \[r=\sqrt{100-c}\]                 Condition of tangency\[\Rightarrow p=r\]                 \[\Rightarrow \left| \frac{2(-8)-(-6)+5}{\sqrt{{{2}^{2}}+{{1}^{2}}}} \right|=\sqrt{100-c}\]                 \[\Rightarrow \sqrt{5}=\sqrt{100-c}\]                 \[\Rightarrow c=95\]