Question

In a right angled triangle, the length of the sides are $a$ and $b(0<a<b)$. A circle passes through the mid-point of the smaller side and touches the hypotenuse at its mid-point.

• A. centre of the circle is $(\frac{2a^2-b^2}{4a},\frac{b}{4})$
• B. centre of the circle is $(\frac{b^2-a^2}{4a},\frac{b}{4})$
• C. radius of the circle is $\frac{b\sqrt{a^2+b^2}}{4a}$
• D. radius of the circle is $\frac{a\sqrt{a^2+b^2}}{4b}$

Answer 1

Answer: (A), (C)

The correct options are
A centre of the circle is $(\frac{2a^2-b^2}{4a},\frac{b}{4})$
C radius of the circle is $\frac{b\sqrt{a^2+b^2}}{4a}$

Let the vertices of the triangle be $O(0,0),A(a,0)$ and $B(0,b)$

Let $C(h,k)$ be the centre and $r$ the radius of the given circle.

Since the circle touches $AB$ at the point $D(\frac{a}{2},\frac{b}{2})$, the centre $C(h,k)$ lies on the line.
$y-\frac{b}{2}=\frac{a}{b}(x-\frac{a}{2})$ or
$2ax-2by=a^2-b^2$
$\therefore 2ah-2bk=a^2-b^2$ .......(1)
Now $r^2=C{D}^2={(h-\frac{a}{2})}^2+{(k-\frac{b}{2})}^2$ ......(2)
Also $r^2={(h-\frac{a}{2})}^2+k^{21}$ .......(3)
(because the circle passes through $(\frac{a}{2},0)$, the mid-point of the smaller side).

From (2) and (3), we get

$k=\frac{b}{4},$and from (1), we get

$h=\frac{2a^2-b^2}{4a}$

Putting these values in (3), we get
$r^2={(\frac{2a^2-b^2}{4a}-\frac{a}{2})}^2+{\left(\frac{b}{4}\right)}^2=\frac{b^4}{16a^2}+\frac{b^2}{16}$
$=\frac{b^4+a^2{b}^2}{16a^2}=\frac{b^2(a^2+b^2)}{(4a{)}^2}$
$\Rightarrow r=\frac{b\sqrt{a^2+b^2}}{4a}$