Question

A circle touches the hypotenuse of a right-angled triangle at its midpoint and passes through the midpoint of the shorter side. If $3$ units and $4$ units are the length of the sides other than hypoteuse and $r$ be the radius of the circle, then the value of $3r$ is

Answer 1

Using coordinate geometry, assume endpoints of hypotenuse as $A\left(\right.3,0\left.\right)$ and $B\left(\right.0,4\left.\right)$ , the third vertex as origin,
such that the midpoint of the hypotenuse is $P\left(\frac{3}{2} , 2\right)$ and the midpoint of the shorter side is $Q\left(\frac{3}{2} , 0\right)$
Let centre be $C \left(\right. h , \, k \left.\right)$
$\because C P ⊥ A B$
$\Rightarrow \frac{2 - k}{\frac{3}{2} - h} = \frac{3}{4}$
$\Rightarrow 6 h - 8 k = - 7$ … (i)
$\because C P = C Q$
$\Rightarrow \left(\right. h - \frac{3}{2} \left.\right)^{2} + \left(\right. k - 2 \left.\right)^{2} = \left(\right. h - \frac{3}{2} \left.\right)^{2} + k^{2}$
$\Rightarrow k = 1$ and
$6 h = 1 \Rightarrow h = \frac{1}{6}$
$\therefore $ radius $\left(\right. r \left.\right) = C Q = \sqrt{\left(\right. \frac{1}{6} - \frac{3}{2} \left.\right)^{2} + 1} = \sqrt{\left(\right. \frac{1 - 9}{6} \left.\right)^{2} + 1} = \frac{5}{3}$
$\therefore 3 r = 5$