Question

The value of $c$ for which the set $\left\{ {\left. {\left( {x,y} \right)} \right|{x^2} + {y^2} + 2x \leqslant 1} \right\} \cap \left\{ {\left. {\left( {x,y} \right)} \right|x - y + c \geqslant 0} \right\}$ contains only one point in common is
$
  {\text{A}}{\text{.}}\left( { - \infty , - 1} \right] \cup \left[ {3,\infty } \right) \\
  {\text{B}}{\text{.}}\left\{ { - 1,3} \right\} \\
  {\text{C}}{\text{.}}\left\{ { - 3} \right\} \\
  {\text{D}}{\text{.}}\left\{ { - 1} \right\} \\
 $

Answer 1

Hint:To solve we have to solve the equation of circle and find radius and centre and then since we need only one point common, the line has to be a tangent to the circle and we find the value of $c$ using distance formula.

Complete step-by-step answer:

$
  {x^2} + {y^2} + 2x \leqslant 1 \\
  {x^2} + {y^2} + 2x - 1 \leqslant 0 \\
 $

Now add $1$ and subtract $1$ on the L.H.S of the equation

$
   \Rightarrow {x^2} + {y^2} + 2x - 1 + 1 - 1 \leqslant 0 \\
   \Rightarrow {\left( {x + 1} \right)^2} + {y^2} \leqslant 2 \\
 $

Centre of this circle is $\left( { - 1,0} \right)$ and radius is $\sqrt 2 $

To contain only one point in common the line given should be a tangent to the circle or the perpendicular distance of line from the centre of the circle should be $\sqrt 2 $.

Now using the formula of perpendicular distance between a line $x - y + c \geqslant 0$ and point $\left( { - 1,0} \right)$, we get

$

  d = \dfrac{{\left| { - 1 - 0 + c} \right|}}{{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} }} = \sqrt 2

\\

  \dfrac{{\left| { - 1 + c} \right|}}{{\sqrt 2 }} = \sqrt 2 \\

   \Rightarrow \left| {c - 1} \right| = 2 \\

   \Rightarrow c - 1 = 2{\text{ or }}c - 1 = - 2 \\

   \Rightarrow c = 3{\text{ or }}c = - 1 \\

  \therefore c = 3, - 1 \\

 $

Therefore, our answer is ${\text{B}}{\text{.}}\left\{ { - 1,3} \right\}$.

Note: The standard equation of circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ where $\left( {h,k} \right)$ is centre and $r$ is the radius. Using this equation, we have to arrive at the centre and radius of the circle in the question.
Also, the perpendicular distance between a line $Ax + By + C = 0$ and a point $\left( {{x_1},{y_1}} \right)$ is as follows:

$d = \dfrac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$ . Now, using these formulas we arrive at our desired answer.