Question

If $3x+y=0$ is a tangent to the circle with centre at the point (2, 1), then the equation of the other tangent to the circle from the origin, is

• A. $9x-13y=0$
• B. $9x+13y=0$
• C. $3x-3=0$
• D. $2x+3=0$

Answer 1

Answer: (B)

$9x+13y=0$

Radius of the circle is equal to the perpendicular distance between $3x+y=0$ and the point (2,1).

Radius $=$ $\frac{3\left(2\right)+1}{\sqrt{(3^2+1^2)}}$ $=$ $\frac{7}{\sqrt{10}}$

Equation of circle is $(x-2{)}^2+(y-1{)}^2$ $=$ $\frac{49}{10}$

Equation of tangent with slope 'm' is $(y-1)$ $=$ $m(x-2)\pm a\sqrt{(1+m^2)}$

On satisfying (0,0) in this equation of the tangent and then squaring, we get the quadratic equation $9m^2+40m+39$ $=$ $0$, roots of which ($m_1,m_2$) will be slopes of tangents passing through origin.

$m_1$ $=$ $-3$

$m_2$ $=$ $\frac{-13}{9}$

Therefore, the required equation is $9y+13x=0$