<s>[INST] Equation of a circle tangent which makes a triangle of area $a^2$ with the coordinate axes What is the equation of the tangent to the circle $x^2+y^2-2ax-2ay+a^2=0$ which makes with the coordinate axes a triangle with area $a^2$?
Attempt: The equation of the circle can be re-written as $(x-a)^2+(y-a)^2=a^2$. Let's say the tangent equation is $\frac{x}{c}+\frac{y}{d}=1$. From the condition that it is a tangent I get 
$|\frac{\frac{a}{c}+\frac{a}{d}-1}{\sqrt{\frac{1}{c^2}+\frac{1}{d^2}}}|=a$. Simplifying I get $a(c+d-\sqrt{c^2+d^2})=cd$ and the area condition gives $\frac{1}{2}cd=a^2$. But I am not able to solve this two equation and get values of $c$ and $d$.
 [/INST] Without loss of generality fix $a=1$. The trick here is to realise that $c$ and $d$ must have different signs, or else the formed triangle is either too big or too small (as pointed out by Jean Marie). With this restriction, the condition that the triangle has area 1 must be modified to
$$-\frac12cd=1\text{ or }cd=-2$$
Substituting this into the other equation we get
$$c+d-\sqrt{c^2+d^2}=-2$$
$$(c+d)+2=\sqrt{(c+d)^2-2cd}=\sqrt{(c+d)^2+4}$$
$$(c+d)^2+4(c+d)+4=(c+d)^2+4$$
$$4(c+d)=0$$
$$c=-d$$
It is easy to see from this and $cd=-2$ that $c=\sqrt2$ and $d=-\sqrt2$, or the other way around. Therefore we have the two lines $\frac x{\sqrt2}-\frac y{\sqrt2}=\pm1$. The solution lines for general $a$ merely multiply the constant on the RHS by $a$:
$$\frac x{\sqrt2}-\frac y{\sqrt2}=\pm a$$
$$x-y=\pm a\sqrt2$$
Here is a plot of the solution lines and the triangles they define.

</s>