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.

The line is outide the circle hance putting the line equation inside the formula for the circle is larger then $r^2$ hence
$$
(x - a)^2 + (mx + c - b)^2 > r^2
$$ l.h.s is minimized for $2(x-a) + 2m(mx + c - b) = 0$ or for $x = (m(b - c) + a)/(m^2+1))$. let $f=m^2+1$, then the expression is minimized by
$$
(m(b - c) + a - af)^2 + (m(m(b - c) + a) + cf - bf)^2 > f^2r^2
$$ or
$$
(m(b - c) - m^2a)^2 + (ma + c - b)^2 > f^2r^2.
$$ Which is
$$
f(ma + c - b)^2 > f^2r^2
$$
$$
(ma + c - b)^2 > fr^2 = (m^2 + 1)r^2
$$
And rearranged gives the formula above.
