Question

The value of ${}^{\prime }{c}^{\prime }$ 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

Answer: (A)

$(-\infty ,-1]\cup [3,\infty )$

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

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

This is a circle. now for line $x-y+c\ge 0$ to have one point in common.

Distance from center of line=radius$=\sqrt{2}$

so,$\mid \frac{-1-0+c}{\sqrt{1^2+(-1{)}^2}}\mid =d=\sqrt{2}$

$⟹\frac{c-1}{\sqrt{2}}$

$⟹c=2+1=3$

and $-c+1=2⟹c=-1$

since circle has region of set inside if

so, $c\le -1$ and $c\ge 3$

thus $cϵ(-\infty ,-1]\cup [3,\infty )$