The equation of the pair of tangents at \((0,1)\) to the circle \(x^{2}+y^{2}-2 x-6 y+6=0\) is

Let \(S \equiv x^{2}+y^{2}-2 x-6 y+6=0\)
and
\(\quad P\left(x_{1}, y_{1}\right)=(0,1) \)
\( S_{1}= x_{1}^{2}+y_{1}^{2}-2 x_{1}-6 y_{1}+6=0 \)
\(=(0)^{2}+(1)^{2}-2(0)-6(1)+6 \)
\(= 1-6+6=1 \)
\( T= x \cdot x_{1}+y \cdot y_{1}-\left(x+x_{1}\right)-3\left(y+y_{1}\right)\)\(+6 \)
\(= x \cdot(0)+y(1)-(x+0)-3(y+1)\)\(+6 \)
\(= 0+y-x-3 y-3+6 \)
\( \quad=-x-2 y+3 \)
\(\begin{aligned} \Rightarrow T^{2} &=(-x-2 y+3)^{2} \\ &=(-x-2 y)^{2}+9+6(-x-2 y) \\ &=x^{2}+4 y^{2}+4 x y+9-6 x-12 y \end{aligned}\)
\(\therefore\) Equation of the pair of tangents \(S \cdot S_{1}=T^{2}\)
\(\left(x^{2}+y^{2}-2 x-6 y+6\right)(1)\)
\(\quad=x^{2}+4 y^{2}+4 x y-6 x-12 y+9\)
\(\Rightarrow 3 y^{2}+4 x y-4 x-6 y+3=0\)