Question

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. $\{-1\}$

Answer 1

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

Given set $\left\{\right(x,y)|x^2+y^2+2x\ge 1\}\cap \left\{\right(x,y)|x-y+c\ge 0\}$

$x^2+y^2+2x\le 1=1$circle of centre $(-1,0)$ radius is $\sqrt{1^2-(-1)}=\sqrt{2}$

$x-y+c\ge 0$ is right side of line we should go towards positive x -axis the conditions gets satisfied one point is common The line is tangent to the circle or perpendicular distance of line $x-y+c=0$ from circle $(-1,0)$ is $\sqrt{2}$

So, $\frac{|-1-0+c|}{\sqrt{1^2+(-1{)}^2}}⟹\frac{|c-1|}{\sqrt{2}}=\sqrt{2}$

$|c-1|=2⟹c=3or-1$ The condition is $x-y+3\ge 0$is satisfied by centre an d that means the circle is on the right side of line $x-y+3=0$

$c=\{-1\}$