In a right angled triangle BC = 5, AB= 4, AC = 3. Let S be the circum circle. Let $S_{1}$ be the circle touching both sides AB and AC and circle S internally. Let $S_{2}$ be the circle touching the produced sides AB and AC of $△^{l e} A B C$ , and touching the circle S externally If $r_{1}, r_{2}$ are radii of circles $S_{1}$ and $S_{2}$ respectively then $r_{1} r_{2} =$

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Detailed Solution

Let A= (0,0), B= (4,0), c= (0,3)
Circumcenter of circle S is (2,3/2) and circumradius = $\frac{5}{2}$
Equation of S is ${(x - 2)}^{2} + {(y - 3 / 2)}^{2} = \frac{25}{4}$ ------------(1)
If circle $S_{1}$ is having radius $r_{1}$ and touching AB and AC
$\Rightarrow$ Its centre is ( $r_{1}, r_{2}$ ).
$S_{1}$ and S touch internally $\Rightarrow \sqrt{{(r_{1} - 2)}^{2} + {(r_{1} - 3 / 2)}^{2}} = | r_{1} - 5 / 2 |$
$\Rightarrow {r_{1}}^{2} - 2 r_{1} = 0 \Rightarrow r_{1} = 2$
If $S_{2}$ is having $r_{2}$ and touching AB and AC $\Rightarrow$ centre of $S_{2} = (r_{1}, r_{2})$
$S_{2}$ touches S externally $\Rightarrow {(r_{2} - 2)}^{2} + {(r_{2} - 3 / 2)}^{2} = {(r_{2} + 5 / 2)}^{2}$
$\Rightarrow {r_{2}}^{2} - 12 r_{2} = 0 \Rightarrow r_{2} = 12 \Rightarrow r_{1} r_{2} = 24$