Discussion Forum

If the slope of the tangent of the circle $S=x^{2}+y^{2}-13=0$ at (2,3)  is m, then the point  $\left(m,\frac{-1}{m}\right)$   is 

Answer:

Explanation:

Given  the equation of circle is 

  $S=x^{2}+y^{2}-13=0$  ...........(i)

On differentiating  it w.r.t x, we get

   $2x+2y \frac{dy}{dx}=0$

 $\Rightarrow$      %$\frac{dy}{dx}=\frac{-x}{y}$

 Now , slope of tangent ,   $m= \frac{dy}{dx}|_{at(2,3)}= \frac{-2}{3}$

  $\Rightarrow$               $m=\frac{-2}{3}$

$\therefore$          $\left(m, -\frac{1}{m}\right)=\left(-\frac{2}{3},\frac{3}{2}\right)$

 On substituting this point in LHS  eq.(i), we get

 LHS= $\frac{4}{9}+\frac{9}{4}-13$

 $=\frac{16+81-468}{36}<0$

$\therefore$      $\left(m, -\frac{1}{m}\right)$   is an internal point with respect to the circle S=0