Question

The shortest distance between the circles ${\left(x-1\right)}^2+{\left(y+2\right)}^2=1$ and ${\left(x+2\right)}^2+{\left(y-2\right)}^2=4$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. $5$

Answer 1

The correct option is B

$2$

Explanation for correct option

We know the general equation of circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ where, $\left(h,k\right)$ is the center and radius $r$ of the circle.

Therefore for the circle ${\left(x-1\right)}^2+{\left(y+2\right)}^2=1$ center $C_2$ is $(1,-2)$ and the radius $B{C}_2$ is $1unit$

For the circle ${\left(x+2\right)}^2+{\left(y-2\right)}^2=4$ center $C_1$is $(-2,2)$ and the radius $A{C}_1$is $2units$

Now, the shortest distance between the circles is $AB=C_1{C}_2-A{C}_1-B{C}_2$

We know that the distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Therefore,

$$\begin{gathered} C_1{C}_2=\sqrt{{(-2-1)}^2+{(2+2)}^2} \\ =\sqrt{9+16} \\ =\sqrt{25} \\ =5units \end{gathered}$$

$\therefore AB=5-1-2=2units$

Thus, the shortest distance between the circles is $2units$,

Hence, option (B) is correct.