Question

Shortest distance between $|x|+|y|=2$ and $x^2+y^2=16$ is

• A. $3$
• B. $2$
• C. $1$
• D. $5$

Answer 1

The correct option is B $2$

Considering the equation $|x|+|y|=2$

$Case1-x\ge 0,y\ge 0$

The considered equation will be

$\Rightarrow x+y=2$

$Case2-x\le 0,y\ge 0$

The considered equation will be

$\Rightarrow -x+y=2$

$Case3-x\le 0,y\le 0$

The considered equation will be

$\Rightarrow -x-y=2$

$Case4-x\ge 0,y\le 0$

The considered equation will be

$\Rightarrow x-y=2$

This means the equation $|x|+|y|=2$ represents a rhombus with vertices at $(2,0);(0,2);(-2,0);(0,-2)$

Also, the equation $x^2+y^2=16$ represents the circle with centre $(0,0)$ and radius $4$

On plotting these two equations on a paper as shown in the attached image, we can clearly see that the minimum distance will be along the axis and is equal to $2$

Hence the correct answer is option $B$