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.