Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):


*

*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^2}+b$ 

*Set the circle and line equations equal to each other as in simultaneous equations.
I was then planning on finding some discriminant and as we know the line and circle do not meet I could set the discriminant < 0 and show it is the same as the long inequality. 
Any solutions/pointers appreciated as I cannot seem to work this problem out.

Substitute the equation for the line into the equation for the circle
\begin{eqnarray*}
(x-a)^2+(mx+c-b)^2= r^2 \\
(m^2+1)x^2 -2x(a+m(b-c))+a^2+(c-b)^2-r^2=0.
\end{eqnarray*}
This is a quadratic in $x$ and for no solutions to exist we require the disciminant to be negative so
\begin{eqnarray*}
(a+m(b-c))^2-(m^2+1)(a^2+(c-b)^2-r^2) < 0
\end{eqnarray*}
Now rearrange a bit to obtain your inequality.
