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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.

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

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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$x=15+7cos2t,y=20+7sin2t$

Use Weierstrass Substitution (https://proofwiki.org/wiki/Weierstrass_Substitution)

and
Minimum value of given expression

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).

$$frac{d(1/f)}{dx}=frac{d}{dx}left(frac{20}x-sqrt{frac{49}{x^2}-left(1-frac{15}xright)^2}right)=0\-frac{20}{x^2}-frac12left(-frac{49cdot2}{x^3}-2left(1-frac{15}xright)left(frac{15}{x^2}right)right)left(frac{49}{x^2}-left(1-frac{15}xright)^2right)^{-1/2}=0\left(frac{49}{x^3}+frac{15}{x^2}-frac{225}{x^3}right)^2=frac{400}{x^4}left(frac{49}{x^2}-1+frac{30}{x}-frac{225}{x^2}right)\left(15x-176right)^2=400left(-x^2+{30}x-176right)\625x^2-17280x+101396=0$$This has solutions $$x_pm=frac{1728}{125}pmfrac{2sqrt{112771}}{125}$$ To maximise $x/y$, we choose the largest value of $x$, giving $x_+=19.197dots$, which corresponds to $y=14.3977dots$. This gives $$frac xy=1.3333dotsapproxfrac43$$

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

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
$$351k^2-600k+176leq0$$ or
$$frac{44}{117}leq kleqfrac{4}{3}.$$
The value $frac{4}{3}$ occurs for $$y=frac{15k+20}{k^2+1}=14.4$$ and $$x=14.4cdotfrac{4}{3}=19.2,$$ which says that $frac{4}{3}$ is a maximal value.