The number of common tangents to the circles $x^2+y^2+4 x-6 y-12=0$ and $x^2+y^2-8 x+10 y+5=0$ is
The number of common tangents to the circles $x^2+y^2+4 x-6 y-12=0$ and $x^2+y^2-8 x+10 y+5=0$ is
4
3
2
1
Solution
Equation of given circles
$
\begin{array}{r}
S_1: x^2+y^2+4 x-6 y-12=0 \\
\text { and } S_2: x^2+y^2-8 x+10 y+5=0
\end{array}
$
$\because$ Centre of $S_1$ is $C_1(-2,3)$ and radius of $S_1$ is $r_1=5$ and centre of $S_2$ is $C_2(4,-5)$ and radius of $S_2$ is $r_2=6$.
$
\begin{aligned}
& \because \quad C_1 C_2=\sqrt{(4+2)^2+(-5-3)^2}=\sqrt{36+64}=10 \\
& \text { and } r_1+r_2=5+6=11 \\
& \because \quad r_1+r_2>c_1 c_2
\end{aligned}
$
$\therefore$ There are only two common tangent to circle $S_1$ and $S_2$.
Hence, option (c) is correct