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 :
$$y=\pm \tfrac{1}{\sqrt{5}}(2x+4).$$

A: Sketch (this answer is to find common tangent lines between the two objects).  If you just want the intersections, substitute $y^2$ from the second equation into the first equation:


*

*First, observe that by implicit differentiation on the hyperbola's equation, we get
$$
\frac{2}{9}x-\frac{1}{2}y\frac{dy}{dx}=0.
$$
Therefore,
$$
\frac{dy}{dx}=\frac{4x}{9y}
$$
Observe that the tangent line is a vertical line when $y=0$; in this case, $x=\pm 3$.  

*On the other hand, completing the square on the circle, we get
$$
(x-4)^2+y^2=16
$$
The vertical tangents for the circle are at $y=0$ and $x=0,8$, so they don't agree with the vertical tangents for the hyperbola, so the circle and hyperbola don't have common vertical tangents.  Throughout the rest of this answer, we assume the tangent line isn't vertical.

*Suppose that $y=mx+b$ is tangent to the circle.  Then when we substitute $y=mx+b$ into the equation for a circle, we get 
$$
x^2-8x+m^2x^2+2mbx+b^2=0.
$$
For this line to be tangent, the discriminant of this quadratic must be zero.  (This happens because otherwise, the line intersects the circle in two positions instead of one point, twice).  In other words,
$$
(2mb-8)^2-4(1+m^2)b^2=0.
$$
Simplifying a bit, we get
$$
-32mb+64-b^2=0.
$$

*If we also try this with the hyperbola, we get
$$
x^2/9-m^2x^2/4-2mbx/4-b^2/4=1
$$
Clearing fractions, we get
$$
4x^2-9m^2x^2-18mbx-9b^2-36=0.
$$
Again, to be tangent, the discriminant must be zero so
$$
(-18mb)^2-4(4-9m^2)(-36-9b^2)=0
$$
Simplifying a bit, we get
$$
576+144b^2-1296m^2=0.
$$

*At this point, we only need to solve the simultaneous system of equations
\begin{align*}
576+144b^2-1296m^2&=0\\
-32mb+64-b^2&=0
\end{align*}
We can make this a little more explicit by writing
$$
32mb=64-b^2
$$
and squaring both sides to get
$$
1024m^2b^2=4096-128b^2+b^4.
$$
Substituting 
$$
\frac{4}{9}+\frac{1}{9}b^2=m^2
$$
we get
$$
1024\left(\frac{4}{9}+\frac{1}{9}b^2\right)b^2=4096-128b^2+b^4
$$
This is quadratic in $b^2$, so let $c=b^2$, use the quadratic formula and work backwards from there.
Please do check the calculations (and make sure to throw out complex solutions).
