Question

If the tangent at $\left(1,7\right)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$, then the value of $c$ is:

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

Answer 1

The correct option is B

$95$

Explanation for the correct option:

Step 1: Find the equation of tangent.

The slope of tangent will be $m=\frac{dy}{dx}$ and the equation of tangent will be $\left(y-y_1\right)=m\left(x-x_1\right)$.

$\begin{array}{rcl}{x}^2& =& y-6\\ \Rightarrow y& =& x^2+6\\ & \Rightarrow & \frac{dy}{dx}=2x\end{array}$

So, ${\left(\frac{dy}{dx}\right)}_{\left(1,7\right)}$ will be $2$ and the equation of the tangent at the point $\left(1,7\right)$ will be:

$\begin{array}{rcl}\left(y-7\right)& =& 2\left(x-1\right)\\ \Rightarrow y-7& =& 2x-2\\ & \Rightarrow & 2x-y+5=0\end{array}$

Step 2: Find the centre of the circle.

$\begin{array}{rcl}{x}^2+y^2+16x+12y+c& =& 0\\ & \Rightarrow & {\left(x+8\right)}^2+{\left(y+6\right)}^2+c-64-36=0\\ \Rightarrow {\left(x+8\right)}^2+{\left(y+6\right)}^2& =& 100-c\end{array}$

So, the centre of the circle is $\left(-8,-6\right)$.

Step 3: Find the value of $c$.

If the tangent touches the circle then, the distance from the centre of the circle is equal to its radius.

Distance of $\left(-8,-6\right)$ from $\begin{array}{rcl}2x-y+5& =& 0\end{array}$ will be

$$\begin{gathered} =\left|\frac{2\left(-8\right)+\left(-1\right)\left(-6\right)+5}{\sqrt{4+1}}\right| \\ =\sqrt{5} \end{gathered}$$

So, $\sqrt{100-c}=\sqrt{5}$

By squaring both sides, we get

$\begin{array}{rcl}100-c& =& 5\\ \Rightarrow c& =& 95\end{array}$

Hence, option B is correct.