Question

The equation of the chord of the circle $a^2+y^2-4x=0$, whose mid point is $(1,0)$ is

• A. $y=2$
• B. $y=1$
• C. $x=2$
• D. $x=1$

Answer 1

Answer: (D)

$x=1$

If $(x_1,y_1)$ is the mid point of the chord of the circle, $x^2+y^2-4x=0$, then its equation is
$x{x}_1+y{y}_1-2(x+x_1)=x_1^2+y_1^2-4x_1$
Put $x_1=1,y_1=0$, we get
$x+0-2(x+1)=1^2+0-4$
$\Rightarrow x=1$