Find maximum value of $\frac xy$ 
If $x^2-30x+y^2-40y+576=0$, find the maximum value of $\dfrac xy$.

First I completed the squares and got $(x-15)^2+(y-20)^2=7^2$, which is the equation of a circle.
I think I need to use some properties but I don't know what to do next.
 A: Using the (7,24,25) Pythagorean triangle in the diagram, I get the tangent point
$$ \left( \frac{96}{5}, \frac{72}{5} \right)  $$
and the maximum of $x/y$ on the circle as $4/3.$

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A: $x=15+7\cos2t,y=20+7\sin2t$
Use Weierstrass Substitution (https://proofwiki.org/wiki/Weierstrass_Substitution)
and 
Minimum value of given expression
A: I used straight up calculus here to get the answer.
Let's start with the fact that the maximum point will have to satisfy the circle equation. So writing 
$$x = 15 + \sqrt{49 - (y-20)^2}$$
We get the ratio as
$$R = \frac{15 + \sqrt{49 - (y-20)^2}}{y}$$
Differentiating this and setting to $0$ would give you a long and harrowing equation to solve which would result in this
$$625y^2 -23040y +202176=0$$
Solving this you'd get two values $y=22.464$ or $y=14.39999$
Checking the corresponding ratios(we ignore the negative values as they will only lower $x$ and hence cannot be a part of maxima). Between the two values and testing random values on the boundary of the circle, we conclude that the maximum is for $y=14.399999$, we get an $x=19.199999$ resulting in a ratio of
$$\frac xy = 1.3333$$
NOTE - I left out a lot of intermediary equation solving which is not difficult, just time consuming and complex. If you want I can add that as well
A: An alternative method, uses calculus. Similar to Sauhard Sharma's answer.

First note, this is the equation of a circle centred on $(15,20)$ of radius $7$, so it does not reach the $x$ axis (otherwise you'd be able to get arbitrarily large values for $x/y$). The value of $x/y$ is obviously going to be maximised on the lower half of the circle, since these have lower values of $y$, with the same values of $x$ as the corresponding points in the upper half of the circle. So we can write $$y=20-\sqrt{49-(x-15)^2}$$
We want to maximise $$f=\frac xy=\frac{x}{20-\sqrt{49-(x-15)^2}}$$
The range of $x$ is $[8,22]$. First we can check if this function has any maxima (or equivalently, if its reciprocal has a minimum).
$$\frac{d(1/f)}{dx}=\frac{d}{dx}\left(\frac{20}x-\sqrt{\frac{49}{x^2}-\left(1-\frac{15}x\right)^2}\right)=0\\-\frac{20}{x^2}-\frac12\left(-\frac{49\cdot2}{x^3}-2\left(1-\frac{15}x\right)\left(\frac{15}{x^2}\right)\right)\left(\frac{49}{x^2}-\left(1-\frac{15}x\right)^2\right)^{-1/2}=0\\\left(\frac{49}{x^3}+\frac{15}{x^2}-\frac{225}{x^3}\right)^2=\frac{400}{x^4}\left(\frac{49}{x^2}-1+\frac{30}{x}-\frac{225}{x^2}\right)\\\left(15x-176\right)^2=400\left(-x^2+{30}x-176\right)\\625x^2-17280x+101396=0$$This has solutions $$x_\pm=\frac{1728}{125}\pm\frac{2\sqrt{112771}}{125}$$ To maximise $x/y$, we choose the largest value of $x$, giving $x_+=19.197\dots$, which corresponds to $y=14.3977\dots$. This gives $$\frac xy=1.3333\dots\approx\frac43$$
A: 1)Circle $(x-15)^2+(y-20)^2=7^2;$
Maximum of $C:= x/y:$
2) Line $y=(1/C)x = mx$ , where $m= 1/C$.
Minimize $m$(m >0 , why?)
There are 2 tangents to the circle that pass through origin.
Distance formula (point to line):
$ \left | \dfrac{-m(15)+ 1(20)}{\sqrt{m^2+1}}\right |=7.$
Quadratic equation in $m$. 
Use the  smaller solution $m$ to get $C_{max} (=1/m).$
https://en.m.wikipedia.org/wiki/Distance_from_a_point_to_a_line
A: Let $\frac{x}{y}=k$.
Thus, $x=ky$ and the equation
$$k^2y^2-30ky+y^2-40y+576=0$$ or
$$(k^2+1)y^2-2(15k+20)y+576=0$$ has real roots, which says that the discriminant must be non-negative.
Id est, $$(15k+20)^2-576(k^2+1)\geq0$$ or
$$351k^2-600k+176\leq0$$ or
$$\frac{44}{117}\leq k\leq\frac{4}{3}.$$
The value $\frac{4}{3}$ occurs for $$y=\frac{15k+20}{k^2+1}=14.4$$ and $$x=14.4\cdot\frac{4}{3}=19.2,$$ which says that $\frac{4}{3}$ is a maximal value.
