How to find range of values for where line cuts circle? I am doing a question on circles. The question is:

The circle  has equation $x^2 + y^2 - 6x + 10y + 9 = 0
$
The line with equation $y= kx$ , where  is a constant, cuts  at two distinct points.
Find the range of values for k

My working so far:
Found centre: (3, -5) radius = 5
Now what do I do after this?
 A: Road map.
0)Draw the problem; By inspection $k=0$ i.e. $y=0$ is a tangent.
Limits of $k$:
1)Tangential distance from line to point (center of circle): $5$
$d=\dfrac{|Ax_{0}+By_{0}+C|}{\sqrt{A^2+B^2}}$ where
$Ax+By+C=0$ is your line, and $(x_0,y_0)$ is the point.
2)This is a quadratic for $k;$ find  $k$. These are the limiting $k's.$
3)Determine the range of $k.$
A: Plugging the line equation in that of the circle,
$$x^2 + k^2x^2 - 6x + 10kx + 9 = 0.$$
This is a quadratic equation that has two distinct roots when
$$(5k-3)^2-9(k^2+1)>0$$
or
$$k\left(k-\frac{15}8\right)>0.$$

$$k<0\text{ or }k>\frac{15}8.$$


Alternatively, the distance of the center to the line must not exceed the radius. For convenience, we use the squared distance, and write
$$\frac{(3k+5)^2}{k^2+1}<25$$
or
$$k\left(k-\frac{15}8\right)>0.$$
A: Hint: Plug the equation of line into circle, you will get a quadratic in 'x'... now you need two distinct points.. what's the condition on discriminant for that?
A: We have two equations:
\begin{align}
(x-3)^2+(y+5)^2 &= 5^2 \tag{1}\label{1} \\[4pt]
y &= kx \, . \tag{2}\label{2}
\end{align}
The points of intersection occur when $x$ and $y$ satisfy both equations $\eqref{1}$ and $\eqref{2}$. We can find these points of intersection by substituting $y=kx$ into $\eqref{1}$:
$$
(x-3)^2+(kx+5)^2=5^2 \, .
$$
This will give us a quadratic equation. Since the line cuts the circle, the quadratic should have two solutions (corresponding to two intersections). So use the discriminant to help you.
A: The key to the problem is to forgo any consideration of a circle or of tangent lines.
Instead, you want to identify the two distinct values of $k$, such that each individual value of $k$ will result in only one value of $(x,y)$ that satisfies the following two constraints:

*

*$x^2 + y^2 - 6x + 10y + 9 = 0$.

*$y = kx$.

The above two constraints result in :
$$x^2 + k^2x^2 - 6x + 10kx + 9 = 0 \implies $$
$$x^2(1 + k^2) + x(10k - 6) + 9 = 0.\tag1$$
This is the critical moment in intuiting the analysis.
In order for equation (1) above to have exactly one solution, its discriminant must equal $(0)$.
Edit
Here is where some fun comes in.  What is the relationship between a specific value of $k$ generating only one solution to equation (1) above, and a specific value of $k$ yielding exactly one point $(x,y)$ that satisfies the two (original) constraints?
Re-stating the question, as the slope of the line $y=kx$ (in effect) goes from $0^\circ$ to $360^\circ$ is it possible that a specific slope would intersect the locus of satisfying points $(x,y)$ in more than one location, where each location had the same value of $(x)$.
Answer
The answer is no because: if a specific value of $k$ only generates one satisfying value of $x$, then there can only be one corresponding satisfying value of $y$, namely $y = kx.$
Thus, you are solving for $k$ such that
$$(10k - 6)^2 - 36(1 + k^2) = 0 \implies $$
$$100k^2 - 120k + 36 - 36 - 36k^2 = 0 \implies $$
$$0 = 64k^2 - 120k = k(64k - 120).\tag2$$
Thus, the desired values for $k$ such that $y = kx$ intersects the circle in exactly one point are
$$\left\{0, \frac{120}{64}\right\}.$$
This is the point where, absent a graph, it is easy to get confused.  Focusing on equation (2) above, you want to identify all values of $k$ such that the discriminant will be positive.  This is because when the discriminant is positive, equation (1) above will have two solutions instead of only one solution.  The pertinent values of $k$ are clearly $k < 0$ or $k > \frac{120}{64}.$
Here is the confusing part.  $k = 0$ corresponds to a horizontal line.  As the slope starts at $0$, moves into a negative slope, and then approaches a vertical slope, $k$ is actually approaching $-\infty.$  Then, when the slope crosses over from a vertical line to a positive slope, $k$ goes from $+\infty$ back to $\frac{120}{64}.$
