Answer
$(a)$ $m = -\frac{12}{5}$
$(b)$ $y = \frac{5}{12}x - \frac{169}{12}$
$(c)$ $m_{x} = \frac{-\sqrt {169-x^2} + 12}{x-5}$
$(d)$ $\frac{5}{12}$, which is the same as the slope of the tangent line found in $(b)$
Work Step by Step
$(a)$ $m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{0-(-12)}{0-5} = -\frac{12}{5}$
$(b)$
Perpendicular means slope is -$\frac{1}{m}=\frac{5}{12}$
$y_{2}-y_{1}=m(x_{2}-x_{1})$
$y-(-12) = \frac{5}{12}(x-5)$
$y = \frac{5}{12}(x-5)-12$
$y = \frac{5}{12}x - \frac{169}{12}$
$(c)$
From $x^2+y^2=169$, $y = -\sqrt {169-x^2}$ (The slope is negative because the tangent line in the fourth quadrant will be decreasing.)
$m_{x} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m_{x} = \frac{-\sqrt {169-x^2}-(-12)}{x-5}$
$m_{x} = \frac{-\sqrt {169-x^2}+12}{x-5}$
$(d)$
Rationalize the denominator to get: $\lim\limits_{x \to 5}\frac{x+5}{\sqrt {169-x^2}+12}$
Plug in $x=5$: $\frac{5+5}{\sqrt {169-5^2}+12}=\frac{5}{12}$
$\lim\limits_{x \to 5}m_{x}=\frac{5}{12}$, which is the same as the slope of the tangent line found in $(b)$
