The value of c for which the set,{(x, y) | x^ | vedclass

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(B) The given circle is $x^2 + y^2 + 2x - 1 = 0$.Completing the square,we get $(x + 1)^2 + y^2 = 2$.This represents a circle with center $(-1, 0)$ and

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$\{-1, 3\}$

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(B) The given circle is $x^2 + y^2 + 2x - 1 = 0$.Completing the square,we get $(x + 1)^2 + y^2 = 2$.This represents a circle with center $(-1, 0)$ and radius $r = \sqrt{2}$.The line is $x - y + c = 0$.For the line to intersect the circle at exactly one point,the perpendicular distance from the center $(-1, 0)$ to the line must be equal to the radius.Distance $d = \frac{|(-1) - (0) + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c - 1|}{\sqrt{2}}$.Setting $d = r$,we have $\frac{|c - 1|}{\sqrt{2}} = \sqrt{2}$.$|c - 1| = 2$.This gives $c - 1 = 2$ or $c - 1 = -2$.Therefore,$c = 3$ or $c = -1$.

The value of $c$ for which the set,$\{(x, y) | x^2 + y^2 + 2x \le 1 \} \cap \{(x, y) | x - y + c \ge 0\}$ contains only one point in common is :

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