Answer
\[
Q(-4,-3) \text { and } P(4,3)
\]
Work Step by Step
(a) The line in two space that passes through the point $P_{0}\left(x_{0}, y_{0}\right)$ and that is parallel to the non-zero vector $\mathbf{v}=\langle a, b\rangle=a \mathbf{i}+b \mathbf{j}$ has parametric equations:
\[
b t+y_{0}=y, \quad a t+x_{0}=x
\]
(b) The line in 3d space that passes through the point $P_{0}\left(x_{0}, y_{0}, z_{0}\right)$ and that is parallel to the non-zero vector $\mathbf{v}=\langle a, b, c\rangle=a \mathbf{i}+b \mathbf{j}+c \mathbf{k}$ has parametric equations
\[
c t+z_{0}=z, \quad b t+y_{0}=y, \quad at+x_{0}=x
\]
The vector equations of these lines can be written as:
\[
\begin{aligned}
\langle x, y\rangle &=\left\langle x_{0}+a t, y_{0}+b t\right\rangle \\
\langle x, y, z\rangle &=\left\langle x_{0}+a t, y_{0}+b t, z_{0}+c t\right\rangle
\end{aligned}
\]
or
\[
\begin{aligned}
&\left\langle x_{0}, y_{0}\right\rangle+t\langle a, b\rangle=\langle x, y\rangle \\
&\left\langle x_{0}, y_{0}, z_{0}\right\rangle+t\langle a, b, c\rangle=\langle x, y, z\rangle
\end{aligned}
\]
From the parametric equation of the line, we get:
\[
\langle 4 t, 3 t\rangle=\langle x, y\rangle \Rightarrow y=3 t \text { and } x=4 t
\]
If we substituted this into the circuit equation, the result would be:
\[
\begin{aligned}
25=y^{2}+x^{2} & \Rightarrow25=(3 t)^{2}+(4 t)^{2} \\
& \Rightarrow 25=9 t^{2}+16 t^{2}\\
& \Rightarrow 1=t^{2} \Rightarrow \pm 1=t_{0}
\end{aligned}
\]
For $\pm 1=t_{0}$, we find that
\[
(\pm 4,\pm 3)=\left(4 t_{0}, 3 t_{0}\right)=\left(x_{0}, y_{0}\right)
\]
So, the line intersects the circle at the points $Q(-4,-3)$, $P(4,3)$
