Question

Find the number of values 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 is common

Answer 1

For only one point to be common the line must be tangent to the given circle

$x^2+y^2+2x-1=0⟹(x+1{)}^2+y^2=2$

Distance of centre of circle $(-1,0)$ from the line is radius $\sqrt{2}$

$\frac{|-1-0+c|}{\sqrt{1+1}}=\sqrt{2}⟹|c-1|=2⟹c=3.-1$

$c$ has two values