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 the radius of circle is

• A. $\frac{b}{a}\sqrt{a^2+b^2}$ units
• B. $\frac{b}{2a}\sqrt{a^2+b^2}$ units
• C. $\frac{b}{4a}\sqrt{a^2+b^2}$ units
• D. None of these

Answer 1

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

The equation of the circle touching $\frac{x}{a}+\frac{y}{b}=1$ at $P$ $(\frac{a}{2},\frac{b}{2})$ is $S+\lambda L=0$ Where $S=0$ is point circle
${(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2+\lambda (\frac{x}{a}+\frac{y}{b}-1)=0\cdots \left(i\right)$
It passes through $Q(\frac{a}{2},0)$
Therefore, $\lambda =\frac{b^2}{2}$
On putting the value of $\lambda$ in Eq. $\left(i\right)$, we get
${(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2+\frac{b^2}{2}(\frac{x}{a}+\frac{y}{b}-1)=0$
$\Rightarrow x^2+y^2-(a-\frac{b^2}{2a})x-\frac{by}{2}+\frac{a^2-b^2}{4}=0$
Let $r$ be the radius of this circle.
Then,
$r^2=\frac{1}{4}{(a-\frac{b^2}{2a})}^2+\frac{b^2}{16}-\left(\frac{a^2-b^2}{4}\right)=\frac{b^2(a^2+b^2)}{16a^2}$
$\Rightarrow r=\frac{b}{4a}\sqrt{a^2+b^2}$ units

Alternate Solution:
Equation of $PC$ will be
$\frac{x}{b}-\frac{y}{a}=\frac{a^2-b^2}{2ab}\cdots \left(1\right)$
Perpendicular bisector of the chord will always pass through the centre of the circle.
Let the coordinate of $C$ will be
$(\alpha ,\frac{b}{4})$
Putting the value of $C$ in $\left(1\right)$
$\alpha =\frac{2a^2-b^2}{4a}$
$\therefore r=\sqrt{{(\frac{2a^2-b^2}{4a}-\frac{a}{2})}^2+\frac{b^2}{16}}$
$\Rightarrow r=\frac{b}{4a}\sqrt{a^2+b^2}$ units