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

It is given that $3x+y=0$ is a tangent to the circle with centre at $(2, -1).$

$\therefore$ Radius = Length of the 1. from $(2, -1) on 3 x + y = 0$

$\Rightarrow$Radius $=\frac{6-1}{\sqrt{9+1}}=\frac{5}{\sqrt{10}}=\sqrt{\frac{5}{2}}$

So, the equation of the circle is 

$(x-2{)}^2+(y+1{)}^2=\frac{5}{2}\Rightarrow x^2+y^2-4x+2y+\frac{5}{2}=0$

The combined equation of the tangents drawn from the origin to this circle is

$\begin{array}{r}\left(x^2+y^2-4x+2y+\frac{5}{2}\right)\left(\frac{5}{2}\right)={\left(-2x+y+\frac{5}{2}\right)}^2\\ \Rightarrow 3x^2-8xy-3y^2=0\Rightarrow 3x+y=0,x-3y=0\end{array}$