Discussion Forum

1)

The number of common tangents to the circles 

$x^{2}+y^{2}-6x-14y+48=0 $  and  $x^{2}+y^{2}-6x=0 $ is 

Answer:

Option D

Explanation:

 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) ;radius r₂=3

 So, r₁+r₂ < d  (distance betweeen the centres)

 $\therefore$  Circle don't  cut and hence the number of common tangents =4