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PQ and PR are two tangents to a circle with centre O and radius $5$ cm. AB is another tangent to the circle at C which lies on OP. If $OP = 13$ cm, then find the length AB and PA.
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Sol. $OP = 13$ cm, $OQ = 5$ cm
$\therefore PQ = \sqrt{169 - 25} = 12$ cm (I Mark)
Let $AC = x = AQ$
$PC = 13 - 5 = 8$ cm and $PA = 12 - x$ (II Mark)
$AC \perp OP \therefore (12 - x)^2 = x^2 + 8^2$ (III Mark)
$\Rightarrow 144 - 24x + x^2 = x^2 + 64$
$\Rightarrow x = \frac{10}{3}$ (IV Mark)
$AB = 2AC = \frac{20}{3}$ cm or $6.6$ cm (approx.) (V Mark)
$PA = 12 - \frac{10}{3} = \frac{26}{3}$ cm or $8.6$ cm (approx.) (VI Mark)
$\therefore PQ = \sqrt{169 - 25} = 12$ cm (I Mark)
Let $AC = x = AQ$
$PC = 13 - 5 = 8$ cm and $PA = 12 - x$ (II Mark)
$AC \perp OP \therefore (12 - x)^2 = x^2 + 8^2$ (III Mark)
$\Rightarrow 144 - 24x + x^2 = x^2 + 64$
$\Rightarrow x = \frac{10}{3}$ (IV Mark)
$AB = 2AC = \frac{20}{3}$ cm or $6.6$ cm (approx.) (V Mark)
$PA = 12 - \frac{10}{3} = \frac{26}{3}$ cm or $8.6$ cm (approx.) (VI Mark)