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 $d$.

The equation of the tangent at $(a+a\cos t,a+a\sin t)$
$$x(\cos t)+y(\sin t)=a(1+\sin t+\cos t)$$
$$a^2=\dfrac{a^2(1+\sin t+\cos t)^2}{2|\sin t\cos t|}$$
$$2+2\sin t+2\cos t+2\sin t\cos t=2|\sin t\cos t|$$
If $\sin t\cos t>0,\sin2t>0,0<2t<\pi$
For $\tan\dfrac t2=u,$
$$1+\sin t+\cos t=0\iff1+\dfrac{2u}{1+u^2}+\dfrac{1-u^2}{1+u^2}=0$$
$u=?$
If $\sin t\cos t<0\iff\sin2t<0,\pi<2t<2\pi\iff\dfrac\pi2< t<\pi$
$$2+2\sin t+2\cos t+2\sin t\cos t=-2\sin t\cos t$$
$$-(\sin t+\cos t)=1+2\sin t\cos t$$
Set $\sin t+\cos t=v,$  $$-v=v^2\implies v=?$$
