Q. The equation of the chord of the circle $x^{2} + y^{2} - 4x = 0$, whose mid point is (1, 0) is
A) $y = 2$
B) $y = 1$
C) $x = 2$
D) $x = 1$
Explanation
(d)
If $\left( x_{1},y_{1} \right)$ is the mid point of the chord of the circle, $x^{2} + y^{2} - 4x = 0$, then its equation is
${xx}_{1} + yy_{1} - 2\left( x + x_{1} \right) = x_{1}^{2} + y_{1}^{2} - 4x_{1}$
Put $x_{1,} = 1,\ y_{1} = 0$, we get
$x + 0 - 2(x + 1) = 1^{2} + 0 - 4\$
$\Rightarrow \ x = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.