Q87 / 160 · TS EAMCET 2017 · Mathematics
If the slope of the tangent of the circle $S \equiv x^2+y^2-13=0$ at $(2,3)$ is $m$, then the point $\left(m, \frac{-1}{m}\right)$ is
- Aan extemal point with respect to the circle $S=0$
- Correct answerBan internal point with respect to the circle $S=0$
- Cthe centre of the circle $S=0$
- Da point on the circle $S=0$
Solution
Given equations of circle is
$$
S \equiv x^2+y^2-13=0
$$
On differentiating it w.r.t $x$, we get
$$
\Rightarrow \quad \begin{aligned}
2 x+2 y \frac{d y}{d x} & =0 \\
\frac{d y}{d x} & =\frac{-x}{y}
\end{aligned}
$$
Now, slope of tangent, $m=\left.\frac{d y}{d x}\right|_{\text {at }(2,3)}=\frac{-2}{3}$
$$
\begin{aligned}
& \Rightarrow \quad m=\frac{-2}{3} \\
& \therefore \quad\left(m,-\frac{1}{m}\right)=\left(-\frac{2}{3}, \frac{3}{2}\right) \\
&
\end{aligned}
$$
On substituting this point in LHS Eq. (i), we get
$$
\begin{aligned}
\text { LHS } & =\frac{4}{9}+\frac{9}{4}-13 \\
& =\frac{16+81-468}{36} < 0
\end{aligned}
$$
$\therefore\left(m,-\frac{1}{m}\right)$ is an internal point with respect to the circle $S=0$.
Hence, the correct answer is option: 'B'.