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Q.
If $(4,0)$ is a point on the circle $x^{2} + ax + y^{2} = 0$, then the centre of the circle is at
Conic Sections
Solution:
$\left(4,0\right)$ 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 $x ^{2}- 4 x + 4 - 4+y^{2}= 0$
$\Rightarrow \left(x-2\right)^{2}+\left(y-0\right)^{2}=4$
$\therefore $ Centre $\left(2,0\right)$ and radius $= 2$