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!
 A: 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
$$ \tan(a+b) = \frac{\tan a+\tan b}{1-\tan a \tan b}= \frac{20+\sqrt{210}}{19}$$
A: 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 $\frac{y_1}{x_1}.$
A: 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
$$
\lambda^* = \frac {1}{19}(20+\sqrt{210})
$$
A: 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.
A: 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:

$$ \tan^{-1}{\dfrac{y_{ tgt}}{x_{ tgt}}}=\tan^{-1}\frac{4}{5}+\tan^{-1}\sqrt{\frac{6}{35}}  $$
Now apply arctangent sum formula and complete the same.
