# When does a line passing through a circle and a set point $P$ have maximum slope?

In the above figure you can see a circumference of radius $$r$$ and a point $$P$$ outside it. Assume that you draw a line going through $$P$$ that hits the circle at some point $$C$$. When does line $$PC$$ have the maximum and minimum slope? Is it when it's tangent to the circle? Why? (Point $$P$$ has to be "under" and "to the left" with respect to the circle).

When I tried to prove it, I thought that it was obvious that the tangent line has more slope than some secant lines (Assume that point P goes clockwise until it gets to the point of tangency). But I do not know what happens after the point of tangency but the secant lines that cut the circle after the point of tangency also cut it before it so then it would be one of those lines that have less slope, but that´s not a rigurous way to prove it.

• Yea, it's going to be one of the two tangents. – Don Thousand Oct 15 '18 at 0:46
• I too want to know the reason. – Vmimi Oct 15 '18 at 0:49
• Extreme Values on a continuous function... – Don Thousand Oct 15 '18 at 0:50
• If the point is "under" the circle, then a tangent won't be the answer. A point that is straight up would create the highest slope. – HugoTeixeira Oct 15 '18 at 1:16
• That´s why I say the location of $P$ is important. Point $P$ has to be "under" and "to the left" with respect to the circle. – Vmimi Oct 15 '18 at 2:00

You can make a function to express express the slope of your line and its point of intersection with the circle. Transform your graph so that you your circle is centered at the x-y axis. This will also shift your point P over maintaining your relationship of maximal slope. Now we can express the slope of the line made with an intersection point on the circle as function with respect to x only $$f(x) = \frac{\sqrt{r^2 - x^2} - P_y}{x-P_x}$$. Take the derivative of this and solve for the max value. So $$\frac{dy}{dx} = \frac{-x*P_y}{\sqrt{r^2-x^2}} - \frac{P_y}{(x-P_x)^2}$$ where x ranges from [-r, r].