Question

The line $y-mx+c$ intersect the circle $x^2+y^2=r^2$ at two points if

• A. $-r\sqrt{1+m^2}<c\le 0$
• B. $0\le c<r\sqrt{1+m^2}$
• C. $-c\sqrt{1-m^2}<r$
• D. $r<c\sqrt{1+m^2}$

Answer 1

Answer: (A), (B)

The correct options are
A $-r\sqrt{1+m^2}<c\le 0$
B $0\le c<r\sqrt{1+m^2}$

The x-coordinates of the points of intersection of the line $y=mx+c$ and the circle $x^2+y^2=r^2$ are given by

$x^2+{(mx+c)}^2=r^2$

$\Rightarrow (1+m^2)x^2+2mcx+c^2-r^2=0$ ...(1)

which, being quadratic in $x$, gives two value of $x$ and hence two points of intersection.

These points will be real distinct is the discriminant of (1) is positive i.e.,

$4m^2{c}^2-4(1+m^2)(c^2-r^2>0)\Rightarrow c^2<r^2(1+m^2)$

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