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

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

Find conditions for lines to be perpendicular

Given, the equation of the line $y=mx+c$

$\Rightarrow \frac{\left(y-mx\right)}{c}=1$

Equation of circle is $x^2+y^2=a^2$

By homogenisation,

$\Rightarrow x^2+y^2–a^2{\left(1\right)}^2=0$

$$\begin{gathered} \Rightarrow x^2+y^2–\frac{a^2{(y–mx)}^2}{c^2}=0 \\ \Rightarrow c^2{x}^2+c^2{y}^2-a^{2}y+a^2{m}^2{x}^2-2ymx{a}^2=0 \end{gathered}$$

For, these lines to be perpendicular coefficient of $x^2$$+$coefficient of $y^2$$=0$

$$\begin{gathered} \Rightarrow (c^2–a^2{m}^2)+(c^2–a^2)=0 \\ \Rightarrow 2c^2=a^2(1+m^2) \end{gathered}$$

Hence, the correct option is $C$, $a^2(m^2+1)=2c^2$.