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The line $y=m x+c$ intercepts the circle $x^2+y^2=r^2$ in two distinct points, if
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Verified Answer
The correct answer is:
$-r \sqrt{1+m^2} < c < r \sqrt{1+m^2}$
Equation of the circle is
$$
x^2+y^2=r^2
$$
and the line is $m x-y+c=0$
The line (ii) intersect (i) in two distinct points, if the length of the $\perp^r$ from the centre $(0,0)$ to the line (ii) is less then $r$
Thus, $\left|\frac{0-0+c}{\sqrt{m^2+1}}\right| < r$
$$
\begin{array}{ccc}
\Rightarrow & |c| < r \sqrt{m^2+1} \\
\Rightarrow & -r \sqrt{m^2+1} < c < r \sqrt{m^2+1}
\end{array}
$$
$$
x^2+y^2=r^2
$$
and the line is $m x-y+c=0$
The line (ii) intersect (i) in two distinct points, if the length of the $\perp^r$ from the centre $(0,0)$ to the line (ii) is less then $r$
Thus, $\left|\frac{0-0+c}{\sqrt{m^2+1}}\right| < r$
$$
\begin{array}{ccc}
\Rightarrow & |c| < r \sqrt{m^2+1} \\
\Rightarrow & -r \sqrt{m^2+1} < c < r \sqrt{m^2+1}
\end{array}
$$
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