Question

The number of common tangents that can be drawn to the circles $x^2+y^2-12x+8y+48=0$ and $x^2+y^2-4x+2y-4=0$ is

• A. $0$
• B. $4$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

$x^2+y^2-12x+8y+48=0$ [ From point (1) ]
$r_1=2$
$C_1\equiv (6,-4)$
and $x^2+y^2-4x+2y-4=0$
$C_2\equiv (2,-1)$
$r_2\equiv 3$
Distance between centre $=\sqrt{4^2+3^2}$
$=5=r_1+r_2$
So, This two circle touch each other.
So. no of common tangent $=2$ external tangents
$+1$ internal tangent
$=3$ tangent common,