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Q.
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
ManipalManipal 2015
Solution:
Since, $3 x+y=0$ is a tangent to the circle with centre at $(2,-1)$.
$\therefore $ Radius $=$ Length of the perpendicular 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
$(x-2)^{2}+(y+1)^{2}=\frac{5}{2}$
$\Rightarrow x^2 +y^{2}-4 x+2 y+\frac{5}{2}=0$
The combined equation of the tangents drawn from the origin to this circle is
$S S_{1}=T^{2}$
where, $S=x^{2}+y^{2}-4 x+2 y+\frac{5}{2}$,
$S_{1}=0^{2}+0^{2}-4 \times 0+2 \times 0+\frac{5}{2}$
$=\frac{5}{2}$
and $T=x(0)+y(0)-2(x+0)+(y+0)+\frac{5}{2}$
$=-2 x+y+\frac{5}{2}$
$=-2 x+y+\frac{5}{2}$
$\therefore \left(x^{2}+y^{2}-4 x+2 y+\frac{5}{2}\right)\left(\frac{5}{2}\right)$
$=\left(-2 x+y+\frac{5}{2}\right)^{2}$
$\Rightarrow 3 x^{2}-8 x y-3 y^{2}=0$
$\Rightarrow 3 x+y=0$ and $x-3 y=0$