#### Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ.

4.8 cm
Solution

Given : Radii of two circles are OP = 3 cm and ${O}'$  and intersection point of two circles are P and Q.  Here two tangents drawn at point P are OP and ${O}'$P
$\therefore \: \angle P= 90^{\circ}$
$3^{2}-x^{2}= \left ( NP \right )^{2}$
Also apply Pythagoras theorem in $\bigtriangleup PN{O}'$  we get
$\left ( P{O}' \right )^{2}= \left ( PN \right )^{2}+\left (N {O}' \right )^{2}$
$\left (4 \right )^{2}= \left ( PN \right )^{2}+\left (5-x \right )^{2}$
$16= \left ( PN \right )^{2}+\left (5-x \right )^{2}$
$16- \left (5-x \right )^{2}= \left ( PN \right )^{2}\cdots \left ( ii \right )$
Equate equation (i) and (ii) we get
$9-x^{2}= 16-\left ( 5-x \right )^{2}$
$9-x^{2}-16+\left ( 25+x^{2}-10x \right )= 0$
$\left [ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right ]$
$-7-x^{2}+\left ( 25+x^{2}-10x \right )= 0$
$-7-x^{2}+25+x^{2}-10x= 0$
$18-10x= 0$
$18= 10x$
$\frac{18}{10}= x$

$x= 1\cdot 8$
Put x = 1.8 in equation (i) we get
$9-\left ( 1\cdot 8 \right )^{2}= NP^{2}$
$9-3\cdot 24= NP^{2}$
$5\cdot 76= NP^{2}$
$NP= \sqrt{5\cdot 76}$
$NP= 2\cdot 4$
$P\mathbb{Q}= 2\times PN= 2\cdot 24= 4\cdot 8 cm$