Question

Find the equation to the circle which touches the axes of coordinates and also the line $\frac{x}{a}+\frac{y}{b}=1$, the centre being in the positive quadrant.

Answer 1

The circle touching the co-ordinate axes can be written as

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

$x^2+y^2-2xr-2yr+r^2=0$

Since the line $bx+ay-ab=0$ is touching the circle,it means it is a tangent to the circle.

So, $r=\frac{|br+ar-ab|}{\sqrt{(a^2+b^2)}}$

Solving we get,

$r=\frac{a+b+\sqrt{a^2+b^2}}{2}$ and $r=\frac{a+b-\sqrt{a^2+b^2}}{2}$

Thus the equation of circle is

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

and ${(x-\frac{a+b-\sqrt{a^2+b^2}}{2})}^2+{(y-\frac{a+b-\sqrt{a^2+b^2}}{2})}^2={\left(\frac{a+b-\sqrt{a^2+b^2}}{2}\right)}^2$