Question

The number of common tangents to the circles $x^2+y^2=4$ and $x^2+y^2-8x+12=0$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

The equation of the circles are

$x^2+y^2=4$ ...(1)

and $x^2+y^2-8x+12=0$ ...(2)

Center of (1) is $C_1\equiv (0,0)$ and radius $r_1=2$

Center of (2) is $C_2\equiv (4,0)$ and radius $r_2=2$

$d=$ distance between centers $=C_1{C}_2=4$.

Since $C_1{C}_2=r_1+r_2$,

$\therefore$ the two circles touch each other externally.

Hence $3$ common tangents can be drawn to the two circles.