Question

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

• A. $-r\sqrt{1+m^2}<c<r\sqrt{1+m^2}$
• B. $-c\sqrt{1-m^2}<r<c\sqrt{1+m^2}$
• C. $-r\sqrt{1-m^2}<c<r\sqrt{1+m^2}$
• D. None of these

Answer 1

Answer: (A)

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

Given line is $y=mx+c$ ...(1)

and the given circle is $x^2+y^2=r^2$ ...(2)

Solving (1) and (2), we get

$(1+m^2)x^2+2mcx+c^2-r^2=0$ ...(3)

For two real distinct points of intersection, both the roots of (3) must be real and distinct

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