Question

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

• A. $(-\infty ,-1]\cup [3,\infty )$
• B. $\{-1,3\}$
• C. $\{-3\}$
• D. $\{-1\}$

Answer 1

The correct option is D $\{-1\}$

$x^2+y^2+2x-1=0$
Centre $(-1,0)$ and radius $=\sqrt{2}$
Line $x-y+c=0$ must be tangent to the circle.
$\Rightarrow \left|\frac{-1+c}{\sqrt{2}}\right|=\sqrt{2}$
$\Rightarrow |c-1|=2$
$\Rightarrow c-1=\pm 2$
$\Rightarrow c=3or-1$
$\Rightarrow c=1$ ($\because$ for $c=3$ there will be infinite point commence lying inside circle)