Question

The circles $(x-1{)}^2+y^2=a^2$ and $(x+2{)}^2+y^2=b^2$ touch externally if

• A. $a-b=3$
• B. $a^2+b^2=1$
• C. $a+b=1$
• D. $a+b=3$

Answer 1

The correct option is B $a^2+b^2=1$

$(x-1{)}^2+y^2=a^2\Rightarrow (x-1{)}^2+y^2=a^2$

$x^2-2x+1+y^2=a^2$

$x^2+y^2+2(-1)x+a^2=0$

$\therefore$ center $\Rightarrow (+1,0)$

radius $\Rightarrow \sqrt{(-1{)}^2+0^2-(1-a^2)}$

radius $=\sqrt{a^2}$

Radius $=a$

$(x+2{)}^2+y^2=b^2$

$x^2+y^2+4x+4-b^2=0$

$x^2+y^2+2\left(2\right)x+4-b^2=0$

Center $\Rightarrow (-2,0)$

Radius $=\sqrt{(2{)}^2+0^2-(4-b^2)}$

Radius $=b$

As the two given circles touch externally, the distance between their centers $C_1$$C_2$ is equal to the sum of their radii.

$i.e.C_1{C}_2=r_1+r_2$

$\Rightarrow -2+1=\sqrt{(0+a{)}^2+(0+b{)}^2}$

$1=a^2+b^2$

$\Rightarrow$ option $B$ is correct