The equation of the circle with centre on the | vedclass

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Question

(C) Let the centre of the circle be $(h, 0)$ since it lies on the $x$-axis.Given the radius $r = 4$ and the circle passes through the origin $(0, 0)$.

Vedclass · Accepted Answer

$x^2 + y^2 \pm 8x = 0$

Vedclass · Answer

(C) Let the centre of the circle be $(h, 0)$ since it lies on the $x$-axis.Given the radius $r = 4$ and the circle passes through the origin $(0, 0)$.The distance from the centre $(h, 0)$ to the origin $(0, 0)$ must be equal to the radius $4$.So,$\sqrt{(h-0)^2 + (0-0)^2} = 4$,which implies $|h| = 4$,so $h = \pm 4$.The centre is $(\pm 4, 0)$.The equation of a circle with centre $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.Substituting $h = \pm 4, k = 0, r = 4$:$(x \mp 4)^2 + (y-0)^2 = 4^2$$x^2 \mp 8x + 16 + y^2 = 16$$x^2 + y^2 \mp 8x = 0$.

The equation of the circle with centre on the $x$-axis,radius $4$ and passing through the origin,is

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