Tangent to the curve `y=x^2+6` at a point `(1,7)` touches the circle `x^2+y^2+16x+12y+c=0 `at a point `Q`, then the coordinates of `Q` are (A) `(-6,-11)` (B) `(-9,-13)` (C) `(-10,-15)` (D) `(-6,-7)`

Options

A. 195
B. 185
C. 85
D. 95

Answer

D

Explanation

Key idea Equation of tangent to the curve
`x^(2)= 4ay "at"(x_(1), y_(1)) is "xx"_(1)=4a((y+y_(1))/(2))`
Tangent to the curve `x^(2)=y-6 "at" (1,7) " is "`
`x = (y+7)/(2)-6`
`rArr 2x-y+5=0" "(i)`
Equation of circle is `x^(2)+y^(2)+16x + 12y+C=0 " Centre"(-8,-6)`
`r=sqrt(8^(2)+6^(2)-c)=sqrt(100-c)`
Since, line 2x - y + 5 = 0 also touches the circle.
`therefore sqrt(100-c)=|(2(-8)-(-6)+5)/(sqrt(2^(2)+1^(2)))|`
`rArr sqrt(100-c)=|(-16+6+5)/(sqrt5)|`
`rArr sqrt(100-c)=|-sqrt5|`
`rArr 100-c=5`
`rArr c= 95`