Q. If the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ lie along diameters of acircle of circumference $10\pi$, then the equation of the circle is
A) $x^{2} + y^{2} - 2x + 2y - 23 = 0$
B) $x^{2} + y^{2} - 2x - 2y - 23 = 0$
C) $x^{2} + y^{2} + 2x + 2y - 23 = 0$
D) $x^{2} + y^{2} + 2x - 2y - 23 = 0$
Explanation
(a)
The intersection point of two given lines is the centre of circle $ie,\ (1,\ - 1)$
Circumference of circle$= 10\pi$ (given)
$\Rightarrow \ 2\pi r = 10\pi\ \Rightarrow r = 5$
$\therefore$ equation of circle having centre $(1,\ - 1)$ and radius 5 is
$(x - 1)^{2} + (y + 1)^{2} = 5^{2}$
$\Rightarrow \ \ \ x^{2} + y^{2} - 2x + 2y - 23 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.