Q. The points $(5,\ - 7)$ lies outside the circle
A) $x^{2} + y^{2} - 8x = 0$
B) $x^{2} + y^{2} - 5x + 7y = 0$
C) $x^{2} + y^{2} - 5x + 7y - 1 = 0$
D) $x^{2} + y^{2} - 8x + 7y - 2 = 0$
Explanation
(a)
Let $S = x^{2} + y^{2} - 8x$
At point $(5,\ - 7)$
$S = 5^{2} + ( - 7)^{2} - 8(5) = 34 > 0$
So, point lies outside the circle
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.