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

(- 2, 0)

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(0, 2)

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(2, 0)

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(1, 0)

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(3, 0)

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Solution

The correct option is C $(2, 0)$
Since, $(4, 0)$ is a point on the circle $x^{2} + a x + y^{2} = 0$

Therefore, $(4)^{2} + a (4) + 0^{2} = 0$

$\Rightarrow 16 + 4 a = 0$

$\Rightarrow a = - 4$

Thus equation of circle becomes $x^{2} - 4 x + y^{2} = 0$ .

Therefore, equation of circle becomes

$x^{2} - 4 x + y^{2} = 0$

Therefore, centre of the circle is $(2, 0)$ .

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If (4,0) is a point on the circle x2+ax+y2=0, then the centre of the circle is at :

The correct option is C (2,0)Since, (4,0) is a point on the circle x2+ax+y2=0Therefore, (4)2+a(4)+02=0⇒16+4a=0⇒a=−4Thus equation of circle becomes x2−4x+y2=0.Therefore, equation of circle becomesx2−4x+y2=0Therefore, centre of the circle is (2,0).

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