Question

Find the equation of circle which cuts x-axis at a distance $+3$ from origin and cuts an intercept at y-axis of length $6$ units.

Answer 1

Let circle touches $x-$axis at point $E$ and cuts $AB$
intercepts at $y-$axis.
According to questions,
Let $C$ is centre of the circle. Draw $CD\perp AB$. Then $CD=OE=3$ and
$AD=\frac{AB}{2}=\frac{6}{2}=3$
In right angled triangle $ACD$,
$C{A}^2=A{D}^2+C{D}^2$
$=3^2+3^2=9+9+=18$
$=2\times 9=3\sqrt{2}$
$\therefore$ Radius of circle $a=CA=3\sqrt{2}$
whereas $CE=CA=3\sqrt{2}$
Thus, centre of circle will be $(3,3\sqrt{2})$
Equation of circle,
$(x-3{)}^2-(y-3\sqrt{2}{)}^2=(3\sqrt{2}{)}^2$
$\Rightarrow x^2+9-6x+y^2+18-6\sqrt{2}=18$
$\Rightarrow x^2+y^2-6x-6\sqrt{2}+9=0$
Similarly circle will be in $IInd,IIIrd$ and $IVth$ quadrant whose centres will be
$(-3,3\sqrt{2}),(-3,-3\sqrt{2}),(3,-3\sqrt{2})$
$x^2+y^2-6x-6\sqrt{2y}+9=0$
$x^2+y^2+6x-6\sqrt{2y}+9=0$
and $x^2+y^2-6x-6\sqrt{2}+9=0$
Thus, total four circles are possible whose equations is
$x^2+y^2\pm 6x\pm 6\sqrt{2y}+9=0$