<s>[INST] 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.
 [/INST] 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 $\sin t,\cos t$
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