Question

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

Answer 1

Let us assume the circle touches the axes at $(a,0)$ and $(0,a)$ and we get the radius to be$|a|$.

We get the centre of the circle as $(a,a)$.

This point lies on the line $x–2y=3$

$\Rightarrow a–2\left(a\right)=3$

$\Rightarrow -a=3$

$\Rightarrow a=–3$

$\Rightarrow$ Centre $=(a,a)=(-3,-3)$

Radius of the circle $\left(r\right)=|-3|=3$

We have circle with centre $(-3,-3)$ and having radius $3$.

We know that the equation of the circle with centre (h, k) and having radius ‘r’ is given by: $(x–h{)}^2+(y–k{)}^2=r^2$

Now by substituting the values in the equation, we get

$(x–(-3){)}^2+(y–(-3){)}^2=3^2$

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

$x^2+6x+9+y^2+6y+9=9\Rightarrow 9x^2+y^2+6x+6y+9=0$

Therefore, the equation of the circle is $x^2+y^2+6x+6y+9=0$.