Question

Find the equation of the circle with centre on the $x$-axis, radius $4$units and passing through the origin.

Answer 1

Let the equation of the required circle be ${(x-h)}^2+{(y-k)}^2=r^2$

It is given that the radius of the circle is $4$ and its centre lies on the

$x$-axis , (i.e) $k=0$ and $r=4$units.

Hence the equation of the circle is ${(x-h)}^2+y^2=4^2=16$

It is also given that the circle passes through the origin $(0,0)$

$\therefore {(0-h)}^2+0^2=16$

$\Rightarrow h^2=16$ or $h=\pm 4$

Case$I$

When $h=-4,$ then the equation of the circle is ${(x+4)}^2+y^2=16$

on expanding we get, $x^2+y^2+8x+16=16$

or $x^2+y^2+8x=0$ is the required equation of the circle.

Case$II$

When $h=4$ then the equation of the circle is ${(x-4)}^2+y^2=16$

on expanding we get, $x^2+y^2-8x+16=16$

or $x^2+y^2-8x=0$ is the required equation of the circle.