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.

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 $$\frac xy=1.3333\dots\approx\frac43$$
