The number of common tangents to the two circles $x^2+y^2-8 x+2 y=0 \quad$ and $x^2+y^2-2 x-16 y+25=0$ is :
The number of common tangents to the two circles $x^2+y^2-8 x+2 y=0 \quad$ and $x^2+y^2-2 x-16 y+25=0$ is :
1
2
3
4
Solution
The equations of circles are $x^2+y^2-8 x+2 y=0$ and $x^2+y^2-2 x-16 y+25=0$
The centre and radius of first circle are $C_1(4,-1)$ and $\sqrt{17}$ respectively. Also the centre and radius of second circle are $C_2(1,8)$ and $\sqrt{40}$ respectively.
$\because \quad C_1 C_2=\sqrt{(1-4)^2+(8+1)^2}$
$=\sqrt{9+81}=\sqrt{90}$
and $r_1+r_2=\sqrt{17}+\sqrt{40}$
$\because \quad C_1 C_2 < r_1+r_2$
$\therefore$ These two circles internally
Thus the number of common tangents is 2 .