Question

Circles with radii 3, 4 and 5 touch each other externally if P is the point of intersection of tangents to these circles at their points of contact. Find the distance of P from the points of contact

• A. $\sqrt{3}$
• B. $\sqrt{5}$
• C. $\sqrt{7}$
• D. None of these
  

Answer 1

Answer: (B)

$\sqrt{5}$

Let $A$, $B$ and $C$ be the centres of the circles respectively.

We know that, the line $AP$ bisects the angle formed by the sector at centre $A$.

Similarly we can say this for $BP$ and $CP$.

Clearly, the point $P$ is the in-centre of the $△ABC$ and therefore,

$r=\frac{△}{s}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

Now,

$2s=7+8+9=24$ that is $s=12$

Thus,

$r=\sqrt{\frac{5\times 4\times 3}{12}}=\sqrt{\frac{60}{12}}=\sqrt{5}$

Hence, the distance is $\sqrt{5}$ units.