The line y = mx + c intersects the circle x^2 | vedclass

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(C) The perpendicular distance from the center of the circle $(0, 0)$ to the line $mx - y + c = 0$ must be less than the radius $r$ for the line to in

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Both $(a)$ and $(b)$

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(C) The perpendicular distance from the center of the circle $(0, 0)$ to the line $mx - y + c = 0$ must be less than the radius $r$ for the line to intersect the circle at two distinct points.The distance $d$ is given by $d = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{1 + m^2}}$.For two distinct points,we require $d This implies $|c| This inequality can also be written as $-r\sqrt{1 + m^2} < c < r\sqrt{1 + m^2}$.

The line $y = mx + c$ intersects the circle $x^2 + y^2 = r^2$ at two real distinct points,if

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