Discussion Forum

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

Answer:

Explanation:

Key Idea Equation of tangent to the curve

x²=4ay  at ( x₁,y₁) is $xx_{1}=4a( \frac{y+y_{1}}{2})$

 Tagent to the curve  $x^{2}=y-6$ at ( 1,7)

           $x=\frac{y+7}{2}-6$

$\Rightarrow$      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.

$\therefore$     $\sqrt{100-c}= \mid \frac{2( -8)-( -6)+5}{\sqrt{2^{2}+1^{2}}}\mid$

$\therefore$     $\sqrt{100-c}= \mid \frac{-16+6+5}{\sqrt{5^{}}}\mid$

$\therefore$    $\sqrt{100-c}= \mid -\sqrt{5}\mid$

$\therefore$    100-c=5 $\therefore$    c=95