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 B $\{-1,3\}$

$x^2+y^2+2x-1\le 0$ $x-y+c\ge 0$

To contain only one point in common the line should be a tangent to the circle.

$\Rightarrow x^2+y^2+2x+1-1-1\le 0$

$\Rightarrow (x+1{)}^2+y^2\le 2$

Centre= $(-1,0)$

Radius= $\sqrt{2}$

$(-1,0)$ $x-y+c\ge 0$

$d=\left|\frac{-1+c}{\sqrt{2}}\right|=\sqrt{2}$

$\Rightarrow |c-1|=2$

$c-1=2$ $c-1=-2$

$c=3$ $c=-1$

$\therefore c=3,-1$