Question

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

• A. An external point with respect to the circle $S=0$
• B. An internal point with respect to the circle $S=0$
• C. The centre of the circle $S=0$
• D. A point on the circle $S=0$

Answer 1

Answer: (B)

An internal point with respect to the circle $S=0$

$x^2+y^2=13$
tangent at $(2,3)$ is $x\left(2\right)+y\left(3\right)-13=0$
$M=$slope$=\frac{-2}{3}$
Therefore, $(m,-\frac{1}{m})=(\frac{-2}{3},\frac{3}{2})$
$\Rightarrow S(\frac{-2}{3},\frac{3}{2})=\frac{4}{9}+\frac{9}{4}-13=\frac{16+81}{36}-13<0$

Thus the given point is an internal point w.r.t $S=0$