Q. The circle $x^{2} + y^{2} - 3x - 4y + 2 = 0$ cuts $x$-axis at
A) $(2,\ 0),\ ( - 3,\ 0)$
B) $(3,\ 0),\ (4,\ 0)$
C) $(1,\ 0),\ ( - 1,\ 0)$
D) $(1,\ 0),\ (2,\ 0)$
Explanation
(d)
Given equation of circle is $x^{2} + y^{2} - 3x - 4y + 2 = 0$ and it cuts the $x$-axis
$\therefore y = 0$
The equation of the circle becomes
$x^{2} + 0 - 3x - 4(0) + 2 = 0$
$\Rightarrow x^{2} - 3x + 2 = 0\ \Rightarrow x^{2} - 2x - x + 2 = 0$
$\Rightarrow (x - 1)(x - 2) = 0\ \Rightarrow x = 1,\ 2$
Therefore, the points are (1, 0), (2, 0)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.