Question

If $(4,0)$ is a point on the circle $x^2+ax+y^2=0$, then the centre of the circle is at

• A. $(-2,2)$
• B. $(0,2)$
• C. $(2,0)$
• D. $(1,0)$

Answer 1

Answer: (C)

$\left(4,0\right)$ is a point on the circle $x^2+ax+y^2=0$
$\therefore 16+4a+0=0$
$\Rightarrow 4a=-16$
$\Rightarrow a=-4$
$\therefore$ Equation of circle is $x^2-4x+y^2=0$
or $x^2-4x+4-4+y^2=0$
$\Rightarrow {\left(x-2\right)}^2+{\left(y-0\right)}^2=4$
$\therefore$ Centre $\left(2,0\right)$ and radius $=2$