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 legs of the triangle then radius of the circle will be

• A. $\frac{a}{2b}\sqrt{a^2+b^2}$
• B. $\frac{b}{2a}\sqrt{a^2+b^2}$
• C. $\frac{a}{4b}\sqrt{a^2+b^2}$
• D. $\frac{b}{4a}\sqrt{a^2+b^2}$

Answer 1

The correct option is D $\frac{b}{4a}\sqrt{a^2+b^2}$

Let the perpendicular sides of the triangle be along the coordinate axes, and the hypotenuse be $\frac{x}{a}+\frac{y}{b}=1$.
The hypotenuese is tangent at mid point $(\frac{a}{2},\frac{b}{2})$
The equation of the the circle is ${(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2+\lambda (\frac{x}{a}+\frac{y}{b}-1)=0$
Let the shortest side be $a$.
The circle passes through mid-point of the shortest side. $\left(\frac{a}{2},0\right)$
$\therefore \frac{b^2}{4}-\frac{\lambda }{2}=0$
$\Rightarrow \lambda =\frac{b^2}{2}$
The required circle is ${(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2+\frac{b^2}{2}(\frac{x}{a}+\frac{y}{b}-1)=0$
$\Rightarrow x^2+y^2+\frac{b^2-2a^2}{2a}x-\frac{b}{2}y+\frac{a^2-b^2}{4}=0$
$\therefore r^2=g^2+f^2-c={\left(\frac{b^2-2a^2}{4a}\right)}^2+{\left(\frac{b}{4}\right)}^2-\frac{a^2-b^2}{4}$
$r=\frac{b}{4a}\sqrt{a^2+b^2}$