Coordinate Geometry of maximum slope on tangent

Find the maximum value of $$y/x$$ if it satisfies $$(x-5)^2+(y-4)^2=6$$.

Geometrically, this is finding the slope of the tangent from the origin to the circle. Other than solving this equation with $$x^2+y^2=35$$, I cannot see any synthetic geometry solution. Thanks!

Let $$a$$ be the angle between the center line (from origin to circle center) and the $$x$$-axis,

$$\tan a = \frac 45$$

Let $$b$$ be the angle between the center line and the tangent line,

$$\tan b= \frac{\sqrt{6}}{\sqrt{4^2+5^2-6}}= \frac{\sqrt{6}}{\sqrt{35}}$$

The maximum value $$y/x$$ is given by

$$\tan(a+b) = \frac{\tan a+\tan b}{1-\tan a \tan b}= \frac{20+\sqrt{210}}{19}$$

• Typo with sin(b), should be √ (6/41) – albert chan Sep 19 at 13:44
• Thanks, will correct – Quanto Sep 19 at 13:45

See Pole and polar.

The polar line of the origin $$-5x-4y+35=0$$ intersects the circle as seen in the image below, giving the tangents from the origin to the circle. Now you get the two points $$(x_1,y_1)=(\frac{175+4\sqrt{210}}{41}, \frac{140-5\sqrt{210}}{41}),(x_2,y_2)=(\frac{175-4\sqrt{210}}{41},\frac{140+5\sqrt{210}}{41})$$. The maximal ratio $$\frac{y}{x}$$ is then $$\frac{y_2}{x_2}={{140+5\,\sqrt{210}}\over{175-4\,\sqrt{210}}}\approx 1.815335618220497\approx \frac{20+\sqrt{210}}{19},$$ the minimal given by $$\frac{y_1}{x_1}.$$

Let a line be $$y = mx$$

$$(x-5)^2 + (mx-4)^2 - 6 = 0$$ $$(m^2+1)x^2 + (-8m-10)x + 35 = 0$$

If the line does not touch the circle, above have no solution.
In other words, discriminant is negative.

Line is a tangent if discriminant is zero

$$(-8m-10)^2 - 4(m^2+1)(35) = 0$$ $$35(m^2+1)-(4m+5)^2 = 0$$ $$19m^2 -40m + 10 = 0$$ $$m = {20 ± \sqrt{210} \over 19}$$

For maximum slope, pick the + sign case.

The line $$y = \lambda x$$ and the circle $$(x-5)^2+(y-4)^2= 6$$ should intersect and $$\max \frac xy =\max \lambda$$ should be located at a tangency point hence solving for $$x$$

$$(x-5)^2+(\lambda x-4)^2= 6$$

we have

$$x = \frac{4 \lambda +5\pm\sqrt{-19 \lambda ^2+40 \lambda -10}}{\lambda ^2+1}$$

but at tangency

$$-19 \lambda ^2+40 \lambda -10 = 0$$

with

$$\lambda^* = \frac {1}{19}(20+\sqrt{210})$$

Locate the circle center and draw the displaced circle with its radius including other sides using Pythagoras thm. Add two angles at max slope tangent point around origin O as shown directly: $$\tan^{-1}{\dfrac{y_{ tgt}}{x_{ tgt}}}=\tan^{-1}\frac{4}{5}+\tan^{-1}\sqrt{\frac{6}{23}}$$

Now apply arctangent sum formula and complete the same.