Common tangent to two circles 
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
  and
  $$(x - 5)^2 + (y - 4)^2 = 4.$$

I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.

Like in my other answer, the equation of any tangent of $$(x-2)^2+y^2=3^2$$ will be $$x\cos t+y\sin t-(2\cos t+3)=0$$
Now if this has to be tangent of $(x-5)^2+(y-4)^2=4$ if the radius $=$ the distance from the center$(5,4)$
$$\dfrac{|5\cos t+4\sin t-(2\cos t+3)|}{\sqrt{\cos^2t+\sin^2t}}=2$$
$$\implies4\sin t+3\cos t-3=\pm2$$
Case $\#1:+\implies4\sin t+3\cos t=5$
Use Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ to solve for $\sin t,\cos t$
Case $\#2:-\implies4\sin t+3\cos t=1$
Use Weierstrass substitution to find $\sin t,\cos t$
