Question

In the given figure, O is the centre of two concentric circles fo radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal to

(a) $5\sqrt{2}$ cm (b) $3\sqrt{5}$ cm (c) $4\sqrt{10}$ cm (d) $5\sqrt{10}$ cm

Answer 1

In ΔOAP is a right angle at A then
$O{P}^2=O{A}^2+A{P}^2$

OA = 5 cm, OB = 3 cm and AP = 12 cm

$O{P}^2=5^2+{12}^2$
$O{P}^2=25+144=169$

OP = 13 cm

In ΔOBP is a right angle at A then
$O{P}^2=O{B}^2+B{P}^2$

$169=9+B{P}^2$

$B{P}^2=160$

BP = 4√10 cm.

∴ length of BP = 4√10 cm.

Correct option is (c) 4√10cm