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10-1.Circle and System of Circles
normal
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 :
A
$(-\infty , -1] \cup [3, \infty )$
B
$\{-1, 3\}$
C
$\{-3\}$
D
$\{- 1 \}$
Solution

$x^2 + y^2 + 2x – 1 = 0;$
centre $(-1, 0)$ and rad. $=\sqrt{2}$
line $x – y + c = 0$;
$\left| {\frac{{ – 1 + c}}{{\sqrt 2 }}} \right| = \sqrt 2$
$|c – 1| = 2; c – 1 = \pm 2$
$\Rightarrow c = 3$ or $-1$
Standard 11
Mathematics