Q97 / 150 · MHT CET 2012 · Mathematics
The equation of the pair of tangents at $(0,1)$ to the circle $x^{2}+y^{2}-2 x-6 y+6=0$ is
- A$3\left(x^{2}-y^{2}\right)+4 x y-4 x-6 y+3=0$
- Correct answerB$3 y^{2}+4 x y-4 x-6 y+3=0$
- C$3 x^{2}+4 x y-4 x-6 y+3=0$
- D$3\left(x^{2}+y^{2}\right)+4 x y-4 x-6 y+3=0$
Solution
Let $S \equiv x^{2}+y^{2}-2 x-6 y+6=0$
and $\begin{aligned} \quad P\left(x_{1}, y_{1}\right)=(0,1) \\ S_{1}=& x_{1}^{2}+y_{1}^{2}-2 x_{1}-6 y_{1}+6=0 \\=&(0)^{2}+(1)^{2}-2(0)-6(1)+6 \\=& 1-6+6=1 \\ T=& x \cdot x_{1}+y \cdot y_{1}-\left(x+x_{1}\right)-3\left(y+y_{1}\right)+6 \\=& x \cdot(0)+y(1)-(x+0)-3(y+1)+6 \\=& 0+y-x-3 y-3+6 \\ \quad=&-x-2 y+3 \end{aligned}$
$\begin{aligned} \Rightarrow T^{2} &=(-x-2 y+3)^{2} \\ &=(-x-2 y)^{2}+9+6(-x-2 y) \\ &=x^{2}+4 y^{2}+4 x y+9-6 x-12 y \end{aligned}$
$\therefore$ Equation of the pair of tangents $S \cdot S_{1}=T^{2}$
$\left(x^{2}+y^{2}-2 x-6 y+6\right)(1)$
$\quad=x^{2}+4 y^{2}+4 x y-6 x-12 y+9$
$\Rightarrow 3 y^{2}+4 x y-4 x-6 y+3=0$
Hence, the correct answer is option: 'B'.