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

Answer 1

Answer: (A)

Since, $3x+y=0$ is a tangent to the circle with centre at $(2,-1)$.
$\therefore$ Radius $=$ Length of the perpendicular from $(2,-1)$ on $3x+y=0$
$\Rightarrow$ 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
$S{S}_1=T^2$
where, $S=x^2+y^2-4x+2y+\frac{5}{2}$,
$S_1=0^2+0^2-4\times 0+2\times 0+\frac{5}{2}$
$=\frac{5}{2}$
and $T=x(0)+y(0)-2(x+0)+(y+0)+\frac{5}{2}$
$=-2x+y+\frac{5}{2}$
$=-2x+y+\frac{5}{2}$
$\therefore \left(x^2+y^2-4x+2y+\frac{5}{2}\right)\left(\frac{5}{2}\right)$
$={\left(-2x+y+\frac{5}{2}\right)}^2$
$\Rightarrow 3x^2-8xy-3y^2=0$
$\Rightarrow 3x+y=0$ and $x-3y=0$