<s>[INST] Why do we assume that this quadratic has one solution? I'm asking about a question about two lines which are tangents to a circle. Most of the question is quite elementary algebra, it's just one stage I can't get my head round. Picture here:


The circle $C$ has equation $(x-6)^2+(y-5)^2=17$. The lines $l_1$ and $l_2$ are each a tangent to the circle and intersect at the point $(0,12)$. Find the equations of $l_1$ and $l_2$ giving your answers in the form $y=mx+c$.

Both lines have equation $y = mx + 12$ where $m$ represents two gradients to be found (both negative).
The circle has equation $(x-6)^2 + (y-5)^2 = 17$.
Combining the knowledge $y = mx + 12$ for both lines and $(x-6)^2 + (y-5)^2 = 17$ produces the quadratic:
$$(1+m^2)x^2 + (14m−12)x + 68 = 0$$
At this point I was confused about what step to take to get $m$ or $x$. Looking at the worked solution, it says "There is one solution so using the discriminant $b^2 − 4ac = 0$..."
From here it's straightforward algebra again, producing another quadratic based on the $b^2 - 4ac$ of the previous quadratic:
$$(14m−12)^2 - 4 x (1+m^2) + 68 = 0$$
etc. until we have $m = -4$ or $-8/19$.
My question is I don't understand how we can tell it's right to  assume $b^2 - 4ac = 0$ and how we can see that's the right step to take in this question.
Obviously this feels intuitively wrong since we know there are two solutions for m. Is the logic that m is a gradient which intersects with the circle once? But if so, how do you see that this is the right equation to decide it only has one solution? (I had assumed before looking at the worked example if I needed to do something more complicated based on the equations of the radii or the knowledge that the two tangents would be equal length from the circle to where they meet or something.)
As you can tell, this question is based on fairly elementary algebra; I'm more concerned about knowing why this is the right step to take here.
Many thanks for any answers.
 [/INST] An approach using geometry:

Let $P = (0,12)$ and the two tangent points be $A, B$. Then $OP^2 = (0-6)^2 + (12-5)^2 = 85$, and the radius $AO^2 = 17$, so $PA^2 = PB^2 = 68$ as tangents to the circle are the same length.
Draw a line connecting $A$ and $B$, and call its intersection with $OP$, $M$. By similarity ($\Delta PAO \sim \Delta AOM$) and using power of a point, $(2k)^2 = (\sqrt{17} + k)(\sqrt{17} - k) \implies 4k^2 = 17 -k^2 \implies k^2 = 17/5$, where $k = AM = BM$.
The line $OP$ has gradient $\frac{5-12}{6-0} = -\frac{7}{6}$, so the equation of the line is $y = -\frac{7}{6}x + 12$. Making a right triangle with $OM$ as the hypotenuse, $\text{run}^2 + \text{rise}^2 = 17/5$, or $x^2 + \left(\frac{7}{6}x\right)^2 = 17/5 \implies x = -\frac{6}{5}$, so $M$ has $x$-coordinate $6 - \frac{6}{5} = \frac{24}{5}$ and $y$-coordinate $\frac{32}{5}$.
Also, since $OP \perp AB$ using another circle theorem, $AB$ has gradient $\frac{6}{7}$ and $AM$ has length $2k$. Drawing another right triangle with hypotenuse $AM$, $x^2 + \left(\frac{6}{7}x\right)^2 = 68/5 \implies x = -\frac{14}{5}$, so $A$ has coordinates $\left(\frac{24}{5} - \frac{14}{5}, \frac{32}{5} - \frac{14}{5} \cdot \frac{6}{7} \right) = (2,4)$. Now $BM$ has the same length in the opposite direction, so $B = \left(\frac{24}{5} + \frac{14}{5}, \frac{32}{5} + \frac{14}{5} \cdot \frac{6}{7} \right) = \left(\frac{38}{5}, \frac{44}{5} \right)$.
Now use point-slope form to find the equation of the lines $PA,PB$, and you will get $-\frac{8}{19}x + 12, -4x + 12$.
Desmos graph is here.
</s>