Finding the coordinates on a circle given it’s part of a tangent 
There are many parts to the question. The first asks you to find an equation for circle $C_2$, given it has a centre $(10,0)$ and a radius of $3$. This answer is $(x-10)^2+y^2=9$.
We are then told that $ABP$ is a tangent to both circles and cuts the $x$ axis at $P$ and asks us to find the coordinates of $P$, which the answer is $(15,0)$.
This is where I am a bit confused. The question says ‘a line through $P$ has gradient $m$. Write down, in terms of $m$, the equation of this line’. I think this answer is just $y=mx-15m$. 
We are then told that this line cuts circle $C_1$ in two places and the $x$-coordinates of these two points satisfy the equation:
$$x^2(1+m^2)-30m^2x+(225m^2-81)=0$$
Hence determine the coordinates of the point $A$. I am not sure how to find these coordinates. 
 A: Let $u$ be the first coordinate of $P$. Then the Intercept Theorem gives $(u-10)/u=3/9$, from where $u=15$.
Pythagoras shows the $AP$ has length $12$.  Now use the fact that for right-angled triangles its doubled area is the product of their catheti as well as the product of the hypotenuse and the height $h$, which here is the second coordinate of $A$.  Hence from $15h=9\cdot12$ we arrive in $h=7.2$.  Now plug that in the equation of $C_1$ to get $5.4$ for the first coordinate.
Even easier.  If $A(x_0,y_0)$ is a point on $x^2+y^2=r^2$, the tangent's equation in $A$ is given by
$xx_0+yy_0=r^2.$
Proof: Excluding the trivial case $y_0=0$ the tangent's slope is $-x_0/y_0$, it follows that its equation is
$$\frac{y-y_0}{x-x_0}=\frac{-x_0}{y_0}\iff
yy_0-y_0^2=x_0^2-xx_0\iff xx_0+yy_0=x_0^2+y_0^2.$$
Now notice that $x_0^2+y_0^2=r^2.$
Obviously the equation holds in the case $y_0=0$, too.
$\quad\square$
We know that $P(15,0)$ is a point of the tangent in $A(x_0,y_0)$; from there 
$$15\cdot x_0+0\cdot y_0=81$$
it follows that $x_0=81/15$.
A: Let the equation of the line be $y = mx + c$. 
We know the distance of this line from two points vis-a-vis the centers of the two circles. 


*

*Distance from $(0,0)$ is $9$: 


$$\frac{c^2}{1+m^2} = 81$$


*Distance from $(10,0)$ is $3$: 


$$\frac{(10m + c)^2}{1+m^2} = 9$$
Solve for $m$ and $c$.
Edit: 
Here I have used the fact that the distance between a point $(x_1,y_1)$ from a line $ax+by+c = 0$ is given by: 
$$d = \left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|$$
A: I suspect that the following is what the problem-writer intended you to do: When the line through $P$ is tangent to the circle, there’s only one intersection, in which case the polynomial $x^2(1+m^2)-30m^2x+(225m^2-81)$ only has one root. Set the discriminant $-36(16m^2-9)$ to zero and solve for $m$. There will be two solutions; choose the negative value, since the line slopes downward. Substitute this value into the equation for $x$ and solve that equation, then compute the corresponding value of $y$ from the equation of the circle.  
However, as Michael Hoppe’s answer points out, since you likely already worked out the equation of the tangent line when you computed $P$, it’s much easier to compute its pole. Even if you hadn’t, after finding $m$ for the tangent line, it’s much less work to compute its pole that to solve all of those additional quadratic equations above.
