Question

The total number of common tangents of $x^2+y^2-6x-8y+9=0$ and $x^2+y^2=1$ is

• A. 4
• B. 2
• C. 3
• D. 1

Answer 1

Answer: (C)

The given equation of circles
$x^2+y^2-6x-8y+9=0$
and $x^2+y^2=1$
Radius and centre of both circles is
$C_1\to (3,4),R_1=\sqrt{9+16-9}=4$
$C_2\to (0,0),R_2=1$
$C_1{C}_2=\sqrt{(3-0{)}^2+(4-0{)}^2}=\sqrt{9+16}=5$
$R_1+R_2=4+1=5$
$\because C_1{C}_2=R_1+R_2$
In this case, two direct tangent are real and distinct while the transverse tangents are coincident.
So, number of comman tangents $=3$