<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] For the circles $$(x - 2)^2 + y^2 = 9\text{ and  } (x - 5)^2 + (y - 4)^2 = 4,$$
using Article $180$ of The elements of coordinate geometry by Loney,
Case $1:$ If $T_2$ is the point dividing internally the line joining the centres 
in the ratio $2:3$  then its coordinates are  $$\left(\frac{2\cdot2+5\cdot3}{2+3},\frac{2\cdot0+4\cdot3}{2+3}\right)=\left(\frac{19}5,\frac{12}5\right)$$
Now, the equation of any line passing through  $T_2\left(\frac{19}5,\frac{12}5\right)$ is $$\frac{y-\frac{12}5}{x-\frac{19}5}=m\implies 5mx-5y+12-19m=0$$ where $m$ is the gradient
If it is the tangent of the circle, the distance of this line from the centre of each circle will be equal to the radius of the respective circle.
$$\implies \left|\frac{5m\cdot 2-5\cdot0+12-19m}{\sqrt{(5m)^2+(-5)^2}}\right|=3\implies |12-9m|=15\sqrt{m^2+1}$$
$$\text{On squaring and re-arrangement,  } 16m^2+24m+9=0\implies m=-\frac34$$
So, there will be only one crossed common tangent
Case $2:$  If $T_1$ is the point dividing externally the line joining the centres 
in the ratio $2:3$  then its coordinates are  $$\left(\frac{2\cdot2-5\cdot3}{2-3},\frac{2\cdot0-4\cdot3}{2-3}\right)=\left(11,12\right)$$
Now, the equation of any line passing through  $T_1\left(11,12 \right)$ is $$\frac{y-12}{x-11}=n\implies nx-y+12-11n=0$$ where $n$ is the gradient
Can you follow the method used in Case $1$ to find the two values of $n,$ which implies there will be two simple common tangents 
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