Find the equation of a tangent to the circle $x^2+y^2=a^2$ which makes a triangle of area $a^2$ with the coordinate axes.

Question

The tangent intercept the x axis at $(h,0)$ and y axis $(0,k)$. The radius of circle is $a$.

Then $$\frac 12 hk=\frac 12 a\sqrt {h^2+k^2}$$ $$h^2k^2=a^2(h^2+k^2)$$ But $\frac 12 hk=a^2$

So $$h^2k^2=4a^4\tag{1}$$

Now $$4a^4=a^2(h^2+k_2)$$ $$4a^2=h^2+k^2\tag{2}$$

That’s all I could solve, how should I solve this?

Answer 1

WLOG the equation of any tangent $$x\cos t+y\sin t=a$$

$$\dfrac x{a\sec t}+\dfrac y{a\csc t}=1$$

We need $$a^2=\dfrac{a^2|\sec t\csc t|}2=\dfrac{a^2}{|\sin2t|}\implies|\sin2t|=1$$

$$\cos2t=0,2t=(2n+1)\dfrac\pi2\implies t=(2n+1)\dfrac\pi4,0\le n<4$$

Answer 2

The equation of tangent AB to the circle $x^2+y^3=a^2$ is $$y=mx\pm a\sqrt{m^2+1},$$ where $A=(a\frac{\sqrt{1+m^2}}{m},0)$ and $B(0, a\sqrt{1+m^2})$ such that OA and CB are the intercepts on x and y axes,repsectively. The are of $\Delta$OAB is $$\left|a^2\frac{1}{2} \frac{1+m^2}{m} \right| =a^2 \implies m^2=1 \implies m= \pm 1.$$ Four such tangents are possible, these are given as $$y=\pm x \pm a \sqrt{2}.$$