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.

Rewrite the equation of the circle subbing in for $y=mx+c$:
\begin{align*}
(x-a)^2+(y-b)^2&=x^2-2ax+a^2+y^2-2by+b^2\\
&=x^2-2ax+a^2+(mx+c)^2-2b(mx+c)+b^2\\
&=x^2-2ax+a^2+m^2x^2+2mcx+c^2-2bmx-2bc+b^2\\
&=(m^2+1)x^2+2(m(c-b)-a)x+(a^2+b^2+c^2-2bc)=r^2\\
\end{align*}
Therefore solve
$$(m^2+1)x^2+2(m(c-b)-a)x+(a^2+b^2+c^2-2bc-r^2)=0$$
giving
$$x=\frac{-2(m(c-b)-a)\pm\sqrt{4(m(c-b)-a)^2-4(m^2+1)(a^2+b^2+c^2-2bc-r^2)}}{2(m^2+1)}$$
So we require
$$(m(c-b)-a)^2-(m^2+1)(a^2+b^2+c^2-2bc-r^2)<0$$
which simplifies as
$$ m^2(r^2- a^2 ) + 2 a m(b -c) + 2 bc-b^2 - c^2+r^2 <0$$
