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 $(-1,3)$

$\text{Solve: Given,}$

$x^2+y^2+2x⩽1$

$\text{=>}{x}^2+y^2+2x-1⩽0$

$\text{It is a equation of circle having}$

$\text{centre at}(-1,0)\text{and radius}=\sqrt{2}$

$\text{and}x-y+c⩾0\text{is line a line}$

$\text{then, we find to solve both equation}$

$\text{for only one solulion}$

$\text{so, the Given line must be tangent}$

$\text{to the circle.}$

$\text{so, the distance from}(-1,0)\text{to}$

the line equal radius of Circle which is $\sqrt{2}$

$\Rightarrow d=\sqrt{2}=\left|\frac{-1-0+c}{\sqrt{2}}\right|$

$=\sqrt{2}\times \sqrt{2}=|c-1|=2$

$=⟩C=3\text{and}c=-1$

$\Rightarrow c\in \{-1,3\}$