Q. If $P_{1},P_{2},P_{3}$ are the perimeter of the three circles $x^{2} + y^{2} + 8x - 6y = 0,\ 4x^{2} + 4y^{2} - 4x - 12y - 186 = 0$ and $x^{2} + y^{2} - 6x + 6y - 9 = 0$ respectively, then
A) $P_{1} < P_{2} < P_{3}$
B) $P_{1} < P_{3} < P_{2}$
C) $P_{3} < P_{2} < P_{1}$
D) $P_{2} < P_{3} < P_{1}$
Explanation
(b)
Let $r_{1},\ r_{2}$ and $r_{3}$ be the radii of the respective circles, then
$r_{1} = \sqrt{( - 4)^{2} + ( - 3)^{2} + 0} = \sqrt{25} = 5$
$r_{2} = \sqrt{\left( \frac{1}{2} \right)^{2} + \left( \frac{3}{2} \right)^{2} + \left( \frac{186}{4} \right)} = \sqrt{49} = 7$
$r_{3} = \sqrt{(3)^{2} + (3)^{2} + 9} = \sqrt{27} = 3\sqrt{3}$
$\therefore\ \ \ P_{1} = 2\pi r_{1} = 10\pi,\ P_{2} = 2\pi r_{2} = 14\pi,\ P_{3} = 2\pi r_{3} = 6\sqrt{3}\pi$
$\therefore\ \ P_{1} < P_{3} < P_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.