Question

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

• A. $ 3\left(x^{2}-y^{2}\right)+4xy-4x-6y+3=0 $
• B. $ 3y^{2}+4xy-4x-6y+3=0 $
• C. $ 3x^{2}+4xy-4x-6y+3=0 $
• D. $ 3\left(x^{2}+y^{2}\right)+4xy-4x-6y+3=0 $

Answer 1

Answer: (B)

Let $S=x^{2}+y^{2}-2 x-6 y+6=0$
and $ 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 $
$=-x-2 y+3 $
$ \Rightarrow T^{2} =(-x-2 y+3)^{2}$
$=(-x-2 y)^{2}+9+6(-x-2 y) $
$=x^{2}+4 y^{2}+4 x y+9-6 x-12 y$
$\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) $
$=x^{2}+4 y^{2}+4 x y-6 x-12 y+9 $
$\Rightarrow 3 y^{2}+4 x y-4 x-6 y+3=0$