Question

The lines joining the origin to the points of intersection of the line $y=mx+c$ and the circle $x^2+y^2=a^2$ will be mutually perpendicular, if

• A. $a^2(m^2+1)=c^2$
• B. $a^2(m^2-1)=c^2$
• C. $a^2(m^2+1)=2c^2$
• D. $a^2(m^2=1)=2c^2$

Answer 1

The correct option is C $a^2(m^2+1)=2c^2$

Making the equation of circle homogenous with the help of line $y=mx+c$, we get
$x^2+y^2-a^2{\left(\frac{y-mx}{c}\right)}^2=0$
$\Rightarrow c^2{x}^2+c^2{y}^2-a^2{y}^2-a^2{m}^2{x}^2+2a^{2}mxy=0$
$\Rightarrow (c^2-a^2{m}^2)x^2+(c^2-a^2)y^2-2a^{2}mxy=0$ ....(i)
Hence, lines represented by Eq (i) are perpendicular, if
$c^2-a^2{m}^2+c^2-a^2=0$

$\Rightarrow 2c^2=a^2(1+m^2)$