Answer
\[
-4+3 t=y, 3+4 t=x
\]
Work Step by Step
(a) The line in two space that passes through the point $P_{0}\left(x_{0}, y_{0}\right)$ and is parallel to the non-zero vector $\mathbf{v}=\langle a, b\rangle=a \mathbf{i}+b \mathbf{j}$ has parametric equations:
\[
\quad y_{0}+b t=y , x_{0}+a t=x
\]
The vector equations for these lines can be written as:
\[
\left\langle x_{0}+a t, y_{0}+b t\right\rangle=\langle x, y\rangle
\]
or
\[
t\langle a, b\rangle+\left\langle x_{0}, y_{0}\right\rangle=\langle x, y\rangle
\]
Notice that the circle $25=y^{2}+x^{2}$ is centered at $O(0,0) .$ Since the line is tangent to the circle at $P(3,-4)=P\left(x_{0}, y_{0}\right),$ it's perpendicular to the vector
\[
\langle 3,-4\rangle=\overrightarrow{O P}
\]
Since $\vec{v}=\langle 4,3\rangle$ is perpendicular to $\overrightarrow{O P}$
\[
\vec{v} \cdot \overrightarrow{O P}=\langle 3,-4\rangle\langle 4,3\rangle=0
\]
The parametric equation for the line can be written as:
\[
\begin{array}{l}
a t+x_{0}=x, b t+y_{0}=y \\
4 t+3=x, 3 t-4=y
\end{array}
\]
