If the slope of the tangent of the circle S e | vedclass

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(B) Given equation of the circle is $S \equiv x^2+y^2-13=0$.On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\

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

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(B) Given equation of the circle is $S \equiv x^2+y^2-13=0$.On differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{y}$.The slope of the tangent $m$ at $(2,3)$ is $m = \left. \frac{dy}{dx} \right|_{(2,3)} = -\frac{2}{3}$.Now,the point $\left(m, -\frac{1}{m}\right)$ becomes $\left(-\frac{2}{3}, \frac{3}{2}\right)$.To check the position of this point with respect to the circle,we substitute it into the expression $S(x, y) = x^2+y^2-13$:$S\left(-\frac{2}{3}, \frac{3}{2}\right) = \left(-\frac{2}{3}\right)^2 + \left(\frac{3}{2}\right)^2 - 13 = \frac{4}{9} + \frac{9}{4} - 13 = \frac{16 + 81 - 468}{36} = \frac{97 - 468}{36} = -\frac{371}{36}$.Since $S\left(-\frac{2}{3}, \frac{3}{2}\right) < 0$,the point lies inside the circle.

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

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