Equation of a circle tangent which makes a triangle of area $a^2$ with the coordinate axes

Question

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 $d$.

Answer 1

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.

Answer 2

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:

• 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 $d>0.$

Answer 3

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 $\sin t+\cos t=v,$ $$-v=v^2\implies v=?$$

Answer 4

Triangles having its three vertices in the first quadrant (and the symmetrical in the third one) should be discarded because they have more or less area than $a^2$ as is easy to check. The larger case is obvious (black color in the figure below); for the smaller case (green color) we have:

Area of smaller triangles $\lt\int_0^aydx\lt a^2$

Hence the only possibilities are triangles in the fourth quadrant (and the symmetrical at the second one).

The answer for the fourth quadrant is unique because when the point of tangency varies continuously from $(a, 0)$ to $(2a, a)$, the area of the triangles involved varies continuously and increasingly from zero to infinity. Hence (by the theorem of intermediate values) there is only one point of tangency for which the area of the triangle is equal to $a^2$.

Because of both, unicity and the evident equality $\frac{ (\sqrt2 a)(\sqrt2 a)}{2}=a^2$, we have the tangent line suitable to fourth quadrant finally is $$ \frac{x}{\sqrt2 a }+\frac{y}{-\sqrt2 a }=1\iff\color{red}{ y=x-a\sqrt2}$$

It is verified that, in fact, this is the tangent line required and that the point $(x_T,y_T)$ of tangency is $$(x_T,y_T)=\left(\frac{2+\sqrt2}{2}\space a,\space \frac{2-\sqrt2}{2}\space a\right)$$