<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] In fact the problem is that there is a $\pm$ that you haven't considered:
$$\tag{1}|\frac{\frac{a}{c}+\frac{a}{d}-1}{\sqrt{\frac{1}{c^2}+\frac{1}{d^2}}}|=a \ \ \ \Leftrightarrow \ \ \ \pm a(c+d-\sqrt{c^2+d^2})=cd.$$
Thus there are 2 cases:


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*First case, with the plus sign in (1): setting $S=c+d$ and $P=cd$, your system of 2 equations with 2 unknowns $c,d$ can be transformed into


$$\cases{a(S + \sqrt{S^2 - 2P}) = P\\ P = 2a^2}$$
from which we deduce 
$$S+\sqrt{S^2-4a^2}=2a \ \Rightarrow \ \sqrt{S^2-4a^2}=2a-S \ \Rightarrow \ S^2-4a^2=(2a-S)^2  \Rightarrow \ S=2a.$$
We are now with the classical issue: determine two values knowing their sum and their product. The solution is obtained by expressing that $c,d$ are roots of the quadratic:
$$X^2-SX+P=0 \ \ \iff \ \ X^2-2aX+2a^2=0 \ \ \iff \ \ (X-a)^2+a^2=0$$
which has no solutions... unless $a=0$ which is not a realistic solution.
In fact, it is graphically very expectable that there cannot be solutions to this problem: look at the figure below: either the area of the triangle is bigger (case OBC) or it is smaller (case OED) that the area of square $OGAH$:



*

*second case, with the minus sign in (1): Up to you, you will find solutions with $c>0$ and $d<0$ or with $c<0$ and $d>0.$

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