If the tangents x+y+k=0 and x+ay+b=0 drawn to | vedclass

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(C) The given circle is $x^2+y^2+2x-2y+1=0$,which can be written as $(x+1)^2+(y-1)^2=1$. The center is $C(-1, 1)$ and the radius $r=1$.Since the tange

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(C) The given circle is $x^2+y^2+2x-2y+1=0$,which can be written as $(x+1)^2+(y-1)^2=1$. The center is $C(-1, 1)$ and the radius $r=1$.Since the tangents $x+y+k=0$ and $x+ay+b=0$ are perpendicular,their slopes $m_1 = -1$ and $m_2 = -1/a$ satisfy $m_1 m_2 = -1$. Thus,$(-1)(-1/a) = -1$,which gives $a = -1$.The distance from the center $(-1, 1)$ to the tangent $x+y+k=0$ is equal to the radius $r=1$:$\frac{|-1+1+k|}{\sqrt{1^2+1^2}} = 1 \Rightarrow |k| = \sqrt{2}$. Since $k > 1$,we have $k = \sqrt{2}$.The distance from the center $(-1, 1)$ to the tangent $x-y+b=0$ is also $r=1$:$\frac{|-1-1+b|}{\sqrt{1^2+(-1)^2}} = 1 \Rightarrow |b-2| = \sqrt{2}$.This gives $b-2 = \sqrt{2}$ or $b-2 = -\sqrt{2}$. Since $b > 1$,we take $b = 2+\sqrt{2}$.Finally,$b-k = (2+\sqrt{2}) - \sqrt{2} = 2$.

If the tangents $x+y+k=0$ and $x+ay+b=0$ drawn to the circle $S \equiv x^2+y^2+2x-2y+1=0$ are perpendicular to each other and $k, b$ are both greater than $1$,then $b-k=$

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