Question

Find the equation to the circle, of radius $a$, which passes through the two points on the axis of $x$ which are at a distance $b$ from the origin.

Answer 1

Given circle passes through two points on $x$ axis at a distance $b$ from the origin . So it passes through $(b,0)$ and $(-b,0)$

Let the centre of the circle be $(h,k)$

$\Rightarrow \sqrt{{(h-b)}^2+{(k-0)}^2}=\sqrt{{(h+b)}^2+{(k-0)}^2}\Rightarrow {(h-b)}^2+{(k-0)}^2={(h+b)}^2+{(k-0)}^2\Rightarrow h=0$

Now radius of given circle is $a$

$\Rightarrow \sqrt{{(h-b)}^2+{(k-0)}^2}=a\Rightarrow {(h-b)}^2+{(k-0)}^2=a^2\Rightarrow {(0-b)}^2+{(k-0)}^2=a^2\Rightarrow k^2=a^2-b^2\Rightarrow k=\pm \sqrt{a^2-b^2}$

So the centre of the circle is $(0,\pm \sqrt{a^2-b^2})$ and radius of circle is $a$ . So its equation is

${(x-0)}^2+{(y-(\pm \sqrt{a^2-b^2}\left)\right)}^2=a^2{x}^2+y^2-2(\pm \sqrt{a^2-b^2})y+a^2-b^2=a^2{x}^2+y^2\mp 2\sqrt{a^2-b^2}y=b^2$