Question

A circle touches the hypotenuse of a right-angled triangle at its middle point and passes through the mid-point of the shorter side. If $a$ and $b(a<b)$ be the length of the sides, then prove that the radius is $\frac{b}{4a}\sqrt{a^2+b^2}$.

Answer 1

Choose the perpendicular sides along axes of co-ordinates so that the hypotenuse is $\frac{x}{a}+\frac{y}{b}=1$
which is a tangent at mid-point $(a/2,b/2)$.
The equation of the circle by $\left(n_1\right)$ is
$\underset{Pointcircle}{{(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2}+\lambda \underset{Tangent}{(\frac{x}{a}+\frac{y}{b}-1)}=0$
It passes through mid-point $Q$ of $OA$ i.e. $(\frac{a}{2},0)$
$\therefore \frac{b^2}{4}-\frac{\lambda }{2}=0$ or $\lambda =\frac{b^2}{2}$
Required circle is
${(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2+\frac{b^2}{2}(\frac{x}{a}+\frac{y}{b}-1)=0$
or $x^2+y^2+x(-a+\frac{b^2}{2a})+y(-b+\frac{b}{2})+\frac{a^2+b^2}{4}-\frac{b^2}{2}=0$
or $x^2+y^2+\frac{b^2-2a^2}{2a}x-\frac{b}{2}y+\frac{a^2-b^2}{4}=0$
$\therefore r^2=g^2+f^2-c$
$={\left(\frac{b^2-2a^2}{4a}\right)}^2+{\left(\frac{b}{4}\right)}^2-\frac{(a^2-b^2)}{4}$
$=\frac{b^4}{16a^2}+\frac{a^2}{4}-\frac{b^2}{4}+\frac{b^2}{16}-\frac{a^2}{4}+\frac{b^2}{4}$
$=\frac{b^4+a^2{b}^2}{16a^2}\therefore r=\frac{b}{4a}\sqrt{a^2+b^2}$.