Question

The straight line $y=mx+c$ cuts the circle $x^2+y^2=a^2$ at real points if

• A. $\sqrt{\left[a^2\right(1+m^2\left)\right]}\le |c|$
• B. $\sqrt{\left[a^2\right(1-m^2\left)\right]}\le |c|$
• C. $\sqrt{\left[a^2\right(1+m^2\left)\right]}\ge |c|$
• D. $\sqrt{\left[a^2\right(1-m^2\left)\right]}\ge |c|$

Answer 1

The correct option is C $\sqrt{\left[a^2\right(1+m^2\left)\right]}\ge |c|$

For real solution

The intersection points are

$x^2+(mx+c{)}^2=a^2$ and ${\left(\frac{y-c}{m}\right)}^2+y^2=a^2$

Solving each

$x^2(1+m^2)+\left(2mc\right)x+(c^2-a^2)=0y^2(1+m^2)-2cy+(c^2-m^2{a}^2)=0$

For real solution, discriminant $\ge 0$

$D_1=(2mc{)}^2-4(1+m^2\left)\right(c^2-a^2)D_2=(2c{)}^2-4(1+m^2)(c^2-m^2{a}^2)$

$D_1\ge 0:$ $4m^2{c}^2-4[c^2-a^2+m^2{c}^2-m^2{a}^2]\ge 04a^2(1+m^2)-4c^2\ge 0a^2(1+m^2)\ge c^2\sqrt{a^2(1+m^2)}\ge |c|$

$D_2\ge 0:$ $4c^2-4[c^2-m^2{a}^2-m^2{c}^2-m^4{a}^2]\ge 0-4\left[\right(m^2+1\left)\right(-m^2{a}^2)+m^2{c}^2]\ge 0-(m^2+1)\left(m^2{a}^2\right)+m^2{c}^2\le 0c^2\le (m^2+1)a^2|c|\le \sqrt{(m^2+1)a^2}$