If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8 x-6 y=0$, $4 x^2+4 y^2-4 x-12 y-186=0$…
If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8 x-6 y=0$, $4 x^2+4 y^2-4 x-12 y-186=0$ and $x^2+y^2-6 x+6 y-9=0$ respectively, then
$P_1 < P_2 < P_3$
$P_1 < P_3 < P_2$
$P_3 < P_2 < P_1$
$P_2 < P_3 < P_1$
Solution
Given circle are $x^2+y^2+8 x-6 y=0$
$
\begin{gathered}
4 x^2+4 y^2-4 x-12 y-186=0 \text { and } \\
x^2+y^2-6 x+6 y-9=0 .
\end{gathered}
$
Let $r_1, r_2$ and $r_3$ be the radius of the respective circle, then
$
\begin{aligned}
& r_1=\sqrt{(-4)^2+(-3)^2+0}=\sqrt{25}=5 \\
& r_2=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2+\left(\frac{186}{4}\right)}=\sqrt{49}=7 \\
& r_3=\sqrt{(3)^2+(3)^2+9}=\sqrt{27}=3 \sqrt{3} \\
\therefore & P_1=2 \pi r_1=10 \pi \\
& P_2=2 \pi r_2=14 \pi \\
\therefore & P_3=2 \pi r_3=6 \sqrt{3} \pi \\
& P_1 < P_3 < P_2
\end{aligned}
$