Question

Find the equation of the circle which touches the coordinate axes and whose centre lies on the line $x-2y=3$.

Answer 1

Let the centre $(a,b)$ & Radius $r$.

Since the circle touches the coordinates axes the distance of the circle from the axes will be same. Thus, $|a|=|b|=r$

$(x-a{)}^2+(y-b{)}^2=r^2$

$(a,b)$ satisfies the eqn $x-2y=3,a-2b=3$

$a=2b+3$

$(x-3-2b{)}^2+(y-b{)}^2=r^2$

$|b|=r,b^2=r^2$

$(x-3-2b{)}^2+(y-b{)}^2=b^2$

$(0,b)$ is on the circle. $(0-3-2b{)}^2+(b-b{)}^2=b^2$

$b=-1,-3$ $a=1,-3$

Hence, the centre of the circle $(1,-1)$ & $(-3,-3)$

$(x-1{)}^2+(b+1{)}^2=1$ &

$(x+3{)}^2+(y+3{)}^2=9$