Answer
$(a)$ $m = \frac{4}{3}$
$(b)$ $y = -\frac{3}{4}x+\frac{25}{4}$
$(c)$ $m_{x} = \frac{\sqrt {25-x^2}-4}{x-3}$
$(d)$ $\lim\limits_{x \to 3}m_{x}=-\frac{3}{4}$, which is the same as the slope of the tangent line found in $(b)$.
Work Step by Step
$(a)$ $m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{0-4}{0-3} = \frac{4}{3}$
$(b)$
Perpendicular means slope is -$\frac{1}{m}=-\frac{3}{4}$
$y_{2}-y_{1}=m(x_{2}-x_{1})$
$y-4 = -\frac{3}{4}(x-3)$
$y = -\frac{3}{4}(x-3)+4$
$y = -\frac{3}{4}x+\frac{25}{4}$
$(c)$
From $x^2+y^2=25$, $y = \sqrt {25-x^2}$ (The slope is negative because the tangent line in the first quadrant will be decreasing.)
$m_{x} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m_{x} = \frac{\sqrt {25-x^2}-4}{x-3}$
$(d)$
Use L'Hopitals to get:-$\frac{1}{2}(25-x^2)^{-1/2}(-2x)$
Plug in $x=3$: -$\frac{1}{2}(25-3^2)^{-1/2}(-2(3))=-\frac{3}{4}$
$\lim\limits_{x \to 3}m_{x}=-\frac{3}{4}$, which is the same as the slope of the tangent line found in $(b)$.
