Question

The number of common tangents to the circles $x^2+y^2=4$ and $x^2+y^2-4x+2y-4=0$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given circles are $x^2+y^2=4$ and $x^2+y^2-4x+2y-4=0$

i.e. $x^2+y^2=4$ and $(x-2{)}^2-4+(y-1{)}^2-1-4=0$

i.e. $x^2+y^2=4$ and $(x-2{)}^2+(y-1{)}^2=9$

Center and radii of $\left(i\right)$ is

$C_1=(0,0),r_1=2$

Center and radii of $\left(ii\right)$ is

$C_2=(2,-1),r_2=3$

Difference between $C_1$ and $C_2=\sqrt{5}$

Now here,

$1=r_2-r_1<\sqrt{5}=C_1{C}_2<5=r_1+r_2$

$r_2-r_1<C_1{C}_2<r_1+r_2$

So, there will be $2$ common tangents.