For the two circles $x^2+y^2=16$ and $x^2+y^2-2 y=0$, there is/are
For the two circles $x^2+y^2=16$ and $x^2+y^2-2 y=0$, there is/are
one pair of common tangents
two pair of common tangents
three pair of common tangents
no common tangent
Solution
Let, $x^2+y^2=16$ or $x^2+y^2=4^2$ radius of circle $r_1=4$, centre $\mathrm{C}_1(0,0)$ we have, $x^2+y^2-2 y=0$
$
\begin{aligned}
\Rightarrow & x^2+\left(y^2-2 y+1\right)-1=0 \text { or } x^2+(y-1)^2 \\
=& 1^2
\end{aligned}
$
Radius 1, centre $\mathrm{C}_2(0,1)$
$
\begin{aligned}
&\left|C_1 C_2\right|=1 \\
&\left|r_2-r_1\right|=|4-1|=3 \\
&\left|C_1 C_2\right| < \left|r_2-r_1\right|
\end{aligned}
$
no common tangents for these two circles