Question

The number of common tangents to the circles $x^2+y^2-6x-14y+48=0$ and $x^2+y^2-6x=0$ is

• A. $1$
• B. $2$
• C. $0$
• D. $4$

Answer 1

Answer: (D)

$4$

For the first circle, centre $=(3,7)$
Radius $r_1=\sqrt{3^2+7^2-48}=\sqrt{10}$
For the second circle, centre $=(3,0)$ and radius $r_2=3$
So, $r_1+r_2<d$ (distance between the centres)
$\therefore$ Circle don't cut and hence the number of common tangents $=4$.