Question

Three circles with radii $3$ cm, $4$ cm and $5$ cm touch each other externally. If $A$ is the point of intersection of tangents to these circles at !heir points of contact. then !he distance of A from !he points of contact is

• A. $\sqrt{3}$
• B. $2$
• C. $\sqrt{5}$
• D. $\sqrt{6}$

Answer 1

The correct option is C $\sqrt{5}$

From the Fig, $16.9$ it is clear that A is the centre of the incircle of the triangle $PQR$ formed by the lines joining the centres of the given circles, We have $PQ=5+4=9.QR=4+3=7,RP=3+5=8$ distance of A from the points of contact is the radius $r$ of the incircIe of triangle $PQR$ From trigonometry $r=\frac{\Delta }{s}$. where $\Delta$ is the area and $2s$ is the perimeter of the triangle $PQR$
$\Delta =\sqrt{12(12-7)(12-8)(12-9)}=\sqrt{12\times 5\times 4\times 3}=12\sqrt{5}$
$s=\left(\frac{1}{2}\right)(9+7+8)=12$.
Hence $r=\sqrt{5}$
Ans: C