Question

The 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 coordinate of $Q$ then the coordinate of $Q$ are.

• A. $(-6,-11)$
• B. $(-9,-13)$
• C. $(-10,-15)$
• D. $(-6,-7)$

Answer 1

Answer: (D)

$(-6,-7)$

The tangent to the curve $y=x^2+6$ at point $(1,7)$ is given by:

$x\left(1\right)+6=\frac{1}{2}(y+7)$

$\Rightarrow$ $2x+12=y+7$

$\Rightarrow$ $2x-y+5=0$ ----- ( 1 )

$\Rightarrow$ $y=2x+5$ ----- ( 2 )

If equation ( 1 ) is the tangent to the circle $x^2+y^2+16x+12y+c=0$

Substituting the value of $y$ from equation ( 2 ),

$\Rightarrow$ $x^2+(2x+5{)}^2+16x+12(2x+5)+c=0$

$\Rightarrow$ $x^2+4x^2+25+20x+16x+24x+60+c=0$

$\Rightarrow$ $5x^2+60x+85+c=0$ ---- ( 3 )

Since the line ( 1 ) touches the given circle, then discriminant of equation ( 3 ) must be zero.

$\therefore$ $(60{)}^2-4\times 5\times (85+c)=0$

$\Rightarrow$ $3600-20\times (85+c)=0$

$\Rightarrow$ $180-85-c=0$

$\Rightarrow$ $c=95$

Thus equation ( 3 ) can be re-written as $5x^2+60x+85+95=0$

$\Rightarrow$ $5x^2+60x+180=0$

$\Rightarrow$ $x^2+12x+36=0$

$\Rightarrow$ $(x+6{)}^2=0$

$\Rightarrow$ $x=-6$

For $x=-6$

$y=2\times (-6)+5$

$\Rightarrow$ $y=-12+5$

$\Rightarrow$ $y=-7$

If the point of contact is $Q,$ then the co-ordinates of $Q$ are $(-6,-7)$