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Two circles with centres $O$ and $O'$ of radii $6$ cm and $8$ 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 $PQ$.
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$OO' = \sqrt{6^2 + 8^2} = 10$ cm
quad $\{OP \perp O'P\}$
Let $OA = x, O'A = 10 - x$
$AP^2 = 36 - x^2$
Also $AP^2 = 64 - (10 - x)^2$
Therefore $36 - x^2 = 64 - (10 - x)^2$
$\Rightarrow 36 - x^2 = 64 - 100 - x^2 + 20 x$
$\Rightarrow x = 3.6$
In $\triangle PAO, AP^2 = 36 - (3.6)^2 = 23.04$
$AP = 4.8$
Length $PQ = 2 \times AP = 9.6$ cm
quad $\{OP \perp O'P\}$
Let $OA = x, O'A = 10 - x$
$AP^2 = 36 - x^2$
Also $AP^2 = 64 - (10 - x)^2$
Therefore $36 - x^2 = 64 - (10 - x)^2$
$\Rightarrow 36 - x^2 = 64 - 100 - x^2 + 20 x$
$\Rightarrow x = 3.6$
In $\triangle PAO, AP^2 = 36 - (3.6)^2 = 23.04$
$AP = 4.8$
Length $PQ = 2 \times AP = 9.6$ cm