$r<a, (x+a)^2 +y^2=r^2, (x-a)^2 +y^2=r^2$ four tangent lines

Question

Find the equation of the four tangent lines which are tangent to both circles, $(x+a)^2 +y^2=r^2, (x-a)^2 +y^2=r^2$ Do not give it in the form that involves trigonometric ratios.
What are the four tangent lines?

Answer 1

The first pair of tangents is trivial: $y=\pm r$

The second pair is in the form of $y=mx$, now

$$(x\pm a)^{2}+m^{2}x^{2}=r^{2}$$

$$(1+m^{2})x^{2} \pm 2ax+(a^{2}-r^{2})=0$$

For tangency,

$$\Delta=0$$ $$(\pm 2a)^{2}-4(1+m^{2})(a^{2}-r^{2})=0$$ $$m=\pm \frac{r}{\sqrt{a^{2}-r^{2}}}$$

Hence the second pair of tangents is:

$$\sqrt{a^{2}-r^{2}} \, y = \pm r x$$

Answer 2

➀At first $y=±r$

tangent equation is $(x_0+a)(x+a)+y_0y=r^2$

tangent line pass through $(x,y)=(0,0)$

$x_0=-a+r^2/a$

$y_0=±r√(1-r^2/a^2)$

finally $rx±y√(a^2-r^2)=0$

➁another solution

tangent point P

$OP=√a^2-r^2$

tangent line angle =$±r/OP$ and through $(0,0)$

so, $y=±rx/√a^2-r^2$