Q:

Equation of common tangent

Equation of common tangent of circle $x^2  + y^2 -8x =0$  and hyperbola with $x^2/9 - y^2/4 =1.$
I tried this but was left with lengthy solutions can u  please explain with easy solutions.

A:

Let us give the name (H) to the hyperbola and (C) to the circle.
The equation of a tangent to a conic in a certain point is given by the rule explained and proved in (http://www.emathzone.com/tutorials/geometry/equation-of-tangent-to-conic.html). Thus

the equation of the tangent to (H) in $(x_0,y_0)$ is : 

$\tag{1}x\underbrace{\dfrac{x_0}{9}}_{u}+y\underbrace{\dfrac{y_0}{4}}_{v}-1=0$

the equation of the tangent to (C) in $(x_1,y_1)$ is :

$$xx_1+yy_1 - 4(x+x_1)=0 \ \ \ \iff \ \ \ \ x(x_1-4)+yy_1-4x_1=0$$
$\tag{2} \iff \ \ x\underbrace{\dfrac{(x_1-4)}{4x_1}}_{u}+y\underbrace{\dfrac{y_1}{4x_1}}_{v}-1=0$
Thus, as $(x_0,y_0) \in$ (H) and  $(x_1,y_1) \in$ (C), taking account that (1) and (2) must be a common tangent, the four unknowns $x_0,y_0,x_1,y_1$ must satisfy the four equations:
$$\begin{cases}\dfrac{x_0^2}{9}-\dfrac{y_0^2}{4} =1&(a)\\x_1^2  + y_1^2 -8x_1 =0&(b)\\ \dfrac{x_0}{9}=\dfrac{x_1-4}{4x_1}&(c)\\\dfrac{y_0}{4}=\dfrac{y_1}{4x_1}&(d)\end{cases}$$
that is easily solved, by extracting $x_0$ and $y_0$ from (c) and (d), plugging their expression into (a) and solving system (a)+(b), giving finally:
$$(x_0,y_0)=(-\tfrac92,\pm\sqrt{5}) \ \ \ \text{and} \ \ \  (x_1,y_1)=(\tfrac43,\pm \tfrac43\sqrt{5})$$
Giving the following equations for the two common tangents :
$$y=\pm \tfrac{1}{\sqrt{5}}(2x+4).$$

