I was working on a geometry question and took a really long winded route to get an answer. A worked solution I found for it used the discriminant but I don't understand how. The question was this:
The circle C has equation $ x^2 + 6x + y^2 -2y = 7 $. The lines L1 and L2 are tangents to the circle at the points P and Q respectively. They intersect at the point R $(0,6)$. Find the equations of lines L1 and L2, giving your answers in the form $ y=mx + b$.
I began by drawing out the diagram, and went down a very long route that involved using Pythagoras several times to find various magnitudes of distances, establishing that the centre of the circle, $(-3,1)$, forms a square with P, Q, and R. From this, I worked out the gradient of the line from the centre to R, found the gradient of the line perpendicular to it, and then finally could workout the co-ordinates of the points P and Q. Only from here was I able to finish it and workout the equations of the tangents at P and Q.
However, this took about 30 minutes, and when I looked at the worked solutions saw the following much more concise method:
The equations of the lines L1 and L2 are both $y = mx + 6$.
Substituting $y = mx + 6$ into the equation of the circle, expanding, and factorising gives:
$(1+m^2)x^2 + (6+10m)x + 17 = 0$
There is one solution, so using the discriminant $b^2 - 4ac = 0$:
$(6+10m)^2 - 4(1+m^2)(17)=0$
Which expands and factorises to give:
$(4m-1)(m+4)=0$ therefore $m = 1/4 $ or $ m = - 4$
I am really not conceptually understanding why we can use the discriminant in this situation, and would really appreciate someone explaining what is actually happening when we say "there is one solution, so using the discriminant..."
Thank you in advance!
It comes down to the fact that a tangent line to a circle (or any conic, for that matter), intersects the circle at exactly one point. So, you’re looking for lines through $R$ that intersect the circle at exactly one point. Substituting into the equation of the circle gives you a quadratic equation in the $x$-coordinates of the intersection points. We want this equation to have exactly one solution, which means that the quadratic must have a double root, and that happens when its discriminant vanishes.
You need to be a little careful about using this method. If there’s only a single solution for $x$, that could also mean that both intersection points have the same $x$-coordinate. That can’t happen here, though, because it would mean that the line is vertical, but you can’t represent a vertical line with an equation of the form $y=mx+b$.