The equation of the pair of tangents at (0,1) | vedclass

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(B) Let $S \equiv x^{2}+y^{2}-2x-6y+6=0$ and $P(x_{1}, y_{1}) = (0,1)$.First,calculate $S_{1}$ at point $P(0,1)$:$S_{1} = (0)^{2}+(1)^{2}-2(0)-6(1)+6

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$3y^{2}+4xy-4x-6y+3=0$

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(B) Let $S \equiv x^{2}+y^{2}-2x-6y+6=0$ and $P(x_{1}, y_{1}) = (0,1)$.First,calculate $S_{1}$ at point $P(0,1)$:$S_{1} = (0)^{2}+(1)^{2}-2(0)-6(1)+6 = 1-6+6 = 1$.Next,find the equation of the chord of contact $T$:$T = x(x_{1}) + y(y_{1}) - (x+x_{1}) - 3(y+y_{1}) + 6$$T = x(0) + y(1) - (x+0) - 3(y+1) + 6$$T = y - x - 3y - 3 + 6 = -x - 2y + 3$.The equation of the pair of tangents is given by $SS_{1} = T^{2}$:$(x^{2}+y^{2}-2x-6y+6)(1) = (-x-2y+3)^{2}$$x^{2}+y^{2}-2x-6y+6 = x^{2}+4y^{2}+9 + 4xy - 6x - 12y$Rearranging the terms:$x^{2}-x^{2} + y^{2}-4y^{2} + 4xy - 2x+6x - 6y+12y + 6-9 = 0$$-3y^{2} + 4xy + 4x + 6y - 3 = 0$Multiplying by $-1$:$3y^{2}-4xy-4x-6y+3=0$ (Note: Re-evaluating expansion: $(-x-2y+3)^{2} = x^{2}+4y^{2}+9+4xy-6x-12y$. The equation simplifies to $3y^{2}+4xy-4x-6y+3=0$).

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

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