Question
The straight line $y=m x+c$ will touch the circle $x^2+y^2=a^2$, if $c=$ _________________ .

Answer

$\pm a \sqrt{1+m^2}$, because
If the line touches the circle, there must be only one point which would satisfy both the equations
Hence,
$x^2+y^2=a^2$
$\begin{array}{l} \Rightarrow x^2+(m x+c)^2=a^2 \\ {[\because \text { putting } y=m x+c] }\end{array}$
$\Rightarrow x^2\left(1+m^2\right)+2 m c x+c^2=a^2$
$\Rightarrow x^2\left(1+m^2\right)+2 m c x+c^2-a^2=0$
The line touches the circle. Hence only one value of $x$ is possible. Therefore,
Discriminant $=0$
$\Rightarrow 4 m^2 c^2-4\left(1+m^2\right)\left(c^2-a^2\right)=0$
$\begin{array}{lrl}\Rightarrow m^2 c^2-c^2-m^2 c^2+a^2+m^2 a^2=0 \\ \Rightarrow a^2+a^2 m^2-c^2 =0 \\ \Rightarrow c^2 =a^2\left(1+m^2\right) \\ \Rightarrow c = \pm a \sqrt{1+m^2} .\end{array}$

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