JEE Main On 08 April 2018 Question 1
Question: 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]
Options:
A) 85
B) 95
C) 195
D) 185
Show Answer
Answer:
Correct Answer: B
Solution:
$ x^{2}=y-6 $ tangent at $ \text{P(1, 7)} $ $ x\text{.1=}( \frac{y+7}{2} )-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 | \frac{2(-8)-(-6)+5}{\sqrt{2^{2}+1^{2}}} |=\sqrt{100-c} $
$ \Rightarrow \sqrt{5}=\sqrt{100-c} $
$ \Rightarrow c=95 $
