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.
 A: The circle and the line do not meet if the distance $d$ of the center of the circle to the line is larger than the radius of the circle.
Knowing that the distance of a point $(x_0,y_0)$ to a line $ax+by+c=0$ is 
$$\frac{\vert ax_0+by_0+c\vert}{\sqrt{a^2+b^2}}$$
You get the result by squaring the formula above (and replacing with the proper variables)
$$\frac{(ma-b+c)^2}{m^2+1}>r^2$$ and reordering the terms.
A: 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.
A: 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
$$ m^2(r^2- a^2 ) + 2 a m(b -c) + 2 bc-b^2 - c^2+r^2 <0$$
A: 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.
