Question

If the equation of one tangent to the circle with centre at (2,-1) from the origin $3x+y=0$ then the equation of the Other tangent through the origin is

• A. 3x-y=0
• B. x+3y=0
• C. x-3y=0
• D. x+2y=0

Answer 1

The correct option is C x-3y=0

As shown in figure, equation of tangent OA is given as,

$3x+y=0$

$\therefore y=-3x$

Compare this equation with slope intercept form of line i.e. $y=mx+c$, we get,

slope of tangent OA = $m_1=-3$

Now, $C(2,-1)$ is center of circle.

Let slope of line OC is $m_2$

Thus, slope of line OC = $m_2=\frac{y_2-y_1}{x_2-x_1}$

$\therefore m_2=\frac{0-(-1)}{0-2}$

$\therefore m_2=-\frac{1}{2}$

Now, let $\angle AOC=\theta$

$\therefore tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\therefore tan\theta =\left|\frac{-3-\left(\frac{-1}{2}\right)}{1+(-3)\left(\frac{-1}{2}\right)}\right|$

$\therefore tan\theta =\left|\frac{-3+\frac{1}{2}}{1+\frac{3}{2}}\right|$

$\therefore tan\theta =\left|\frac{\frac{-5}{2}}{\frac{5}{2}}\right|$

$\therefore tan\theta =|-1|=1$

$\therefore \theta ={tan}^{-1}\left(1\right)$

$\therefore \theta =\frac{\pi }{4}$

Now, $\angle AOB=2\theta$

$\therefore \angle AOB=2\times \frac{\pi }{4}$

$\therefore \angle AOB=\frac{\pi }{2}$

Thus, tangent OB and tangent OA are perpendicular to each other.

Let $m_3$ = slope of tangent OB.

$\therefore m_1\times m_3=-1$

$\therefore -3\times m_3=-1$

$\therefore m_3=\frac{1}{3}$

Thus, equation of tangent OB passing through origin is,

$y-0=m_3(x-0)$

$\therefore y=\frac{1}{3}x$

$\therefore 3y=x$

$\therefore x-3y=0$

Thus, answer is option (C)