Answer
a) $3x-4y-25=0$
b) $(-3,4)$
Work Step by Step
a) First determine the slope of the line passing through the points $O(0,0)$ and $P(3,-4)$:
$$m_{OP}=\dfrac{y_P-y_O}{x_P-x_O}=\dfrac{-4-0}{3-0}=-\dfrac{4}{3}.$$
Then determine the slope $m$ of the tangent passing through $P$:
$$\begin{align*}
m\cdot m_{OP}&=-1\\
m\cdot \left(-\dfrac{4}{3}\right)&=-1\\
m&=\dfrac{3}{4}.
\end{align*}$$
The equation of the tangent passing through $P$ is:
$$\begin{align*}
y-y_P=m(x-x_P)\\
y-(-4)&=\dfrac{3}{4}(x-3)\\
y&=\dfrac{3}{4}x-\dfrac{9}{4}-4\\
y&=\dfrac{3}{4}x-\dfrac{25}{4}\\
3x-4y-25&=0.
\end{align*}$$
b) Let $Q$ be another point on the circle where the tangent is parallel to the tangent passing through $P$. It follows that the two tangents have the same slope $m=3/4$. $OQ$ is perpendicular to that tangent, so its slope is $-4/3$. But $OP$ has the same slope $-4/3$, so the points $P,O,Q$ are collinear.
It means that the point $Q$ is the symmetrical of point $P$ about the origin, so its coordinates are $Q(-3,4)$.
