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

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Question

(A) The equation of the circle is $x^2+y^2=r^2$ with center $(0,0)$ and radius $r$. The equation of the line is $mx-y+c=0$. The line intersects the ci

Vedclass · Accepted Answer

$-r \sqrt{1+m^2} < c < r \sqrt{1+m^2}$

Vedclass · Answer

(A) The equation of the circle is $x^2+y^2=r^2$ with center $(0,0)$ and radius $r$. The equation of the line is $mx-y+c=0$. The line intersects the circle at two distinct points if the perpendicular distance from the center $(0,0)$ to the line is less than the radius $r$. The perpendicular distance $d$ is given by: $d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2+1}}$. Setting $d $\frac{|c|}{\sqrt{m^2+1}} $|c| This inequality implies: $-r \sqrt{m^2+1} < c < r \sqrt{m^2+1}$.

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

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