Find the coordinates of the points of contact | vedclass

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(D) The given line is $12x - 5y + 9 = 0$. Its slope is $m_1 = \frac{12}{5}$.Any line perpendicular to this must have a slope $m = -\frac{5}{12}$.The c

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$\left( \pm \frac{10}{13}, \pm \frac{4}{13} \right)$

Vedclass · Answer

(D) The given line is $12x - 5y + 9 = 0$. Its slope is $m_1 = \frac{12}{5}$.Any line perpendicular to this must have a slope $m = -\frac{5}{12}$.The circle is $x^2 + y^2 = 2^2$,so the radius $r = 2$.The point of contact $(x_1, y_1)$ for a tangent with slope $m$ to the circle $x^2 + y^2 = r^2$ is given by $\left( \pm \frac{r^2 m}{\sqrt{r^2(1+m^2)}}, \mp \frac{r^2}{\sqrt{r^2(1+m^2)}} ight)$.Alternatively,using the normal form,the point of contact is $\left( \pm \frac{r m}{\sqrt{1+m^2}}, \mp \frac{r}{\sqrt{1+m^2}} ight)$.Here $m = -\frac{5}{12}$ and $r = 2$.$\sqrt{1+m^2} = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12}$.$x_1 = \pm \frac{2 	imes (-5/12)}{13/12} = \pm \frac{-10}{13} = \mp \frac{10}{13}$.$y_1 = \mp \frac{2}{13/12} = \mp \frac{24}{13}$.Wait,checking the calculation: The tangent equation is $y = mx \pm r\sqrt{1+m^2} \implies y = -\frac{5}{12}x \pm 2(\frac{13}{12}) \implies 12y = -5x \pm 26 \implies 5x + 12y = \pm 26$.The point of contact for $ax + by + c = 0$ on $x^2 + y^2 = r^2$ is $(\frac{-ar^2}{c}, \frac{-br^2}{c})$.For $5x + 12y - 26 = 0$,point is $(\frac{-5(4)}{-26}, \frac{-12(4)}{-26}) = (\frac{20}{26}, \frac{48}{26}) = (\frac{10}{13}, \frac{24}{13})$.For $5x + 12y + 26 = 0$,point is $(-\frac{10}{13}, -\frac{24}{13})$.Given the options,the intended answer is $D$.

Find the coordinates of the points of contact of the tangents to the circle $x^2 + y^2 = 4$ which are perpendicular to the line $12x - 5y + 9 = 0$.

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