Question

If the lines joining the origin to the intersection of the line $y=mx+2$ and the circle $x^2+y^2=1$ are at right angles, then:

• A. $m=\sqrt{3}$
• B. $m=\pm \sqrt{7}$
• C. $m=1$
• D. $m=\sqrt{5}$

Answer 1

The correct option is B

$m=\pm \sqrt{7}$

Explanation for the correct option

Given the lines joining the origin to the intersection of the line $y=mx+2$ and the circle $x^2+y^2=1$ are at right angles.

The equation of the lines joining origin and point of intersection of the line given and the circle will be

$\Rightarrow x^2+y^2={\left(\frac{y-mx}{2}\right)}^2$

$\Rightarrow$ $4\left(x^2+y^2\right)=y^2+m^2{x}^2-2mxy$

$\Rightarrow$$x^2(4-m^2)+y^2(4-1)+2mxy=0$

$\Rightarrow x^2(4-m^2)+2mxy+3y^2=0$

Now since these lines are at right angles, so, $\frac{\mathrm{coefficient}\mathrm{of}{\mathrm{x}}^2}{\mathrm{coefficient}\mathrm{of}{\mathrm{y}}^2}=-1$

$$\begin{gathered} \Rightarrow \frac{4-m^2}{3}=-1 \\ \Rightarrow 4-m^2=-3 \\ \Rightarrow 4-m^2+3=0 \\ \Rightarrow m^2=7 \\ \Rightarrow m=\pm \sqrt{7} \end{gathered}$$

Hence, option (B) i.e. $m=\pm \sqrt{7}$ is the correct answer.