Question

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

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

Answer: (D)

Since, centre and radius of a circle $x^2+y^2=4$ are
$C_1(0,0)$ and $2$ respectively
and centre and radius of another circle
$x^2+y^2-8x+12=0$ are
$C_2(4,0)$ and $2$ respectively.
Now, $C_1{C}_2=\sqrt{(4-0{)}^2+0=4}$
and $r_1+r_2=2+2=4$
$\because C_1{C}_2=r_1+r_2$
$\therefore$ Two circles touch each other externally,
so the number of common tangents is $3$.
Note: (i) If $C_1{C}_2=r_1-r_2$ , one tangents are possible.
If $C_1{C}_2>r_1+r_2$ , four tangents are possible.