Question

The equation of the pair of tangents at $(0,1)$ to the circle $x^2+y^2-2x-6y+6=0$ 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 $S=x^2+y^2-2x-6y+6=0$
and $P\left(x_1,y_1\right)=(0,1)$
$S_1=x_1^2+y_1^2-2x_1-6y_1+6=0$
$=(0{)}^2+(1{)}^2-2(0)-6(1)+6$
$=1-6+6=1$
$T=x\cdot x_1+y\cdot y_1-\left(x+x_1\right)-3\left(y+y_1\right)+6$
$=x\cdot (0)+y(1)-(x+0)-3(y+1)+6$
$=0+y-x-3y-3+6$
$=-x-2y+3$
$\Rightarrow T^2=(-x-2y+3{)}^2$
$=(-x-2y{)}^2+9+6(-x-2y)$
$=x^2+4y^2+4xy+9-6x-12y$
$\therefore$ Equation of the pair of tangents
$S\cdot S_1=T^2$
$\left(x^2+y^2-2x-6y+6\right)(1)$
$=x^2+4y^2+4xy-6x-12y+9$
$\Rightarrow 3y^2+4xy-4x-6y+3=0$