Question

Two circles each of radius 2 and centres O and P touch each other as shown in the figure. If AD and
BD are tangents, then the length of BD is​

Answer 1

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Two circles each of radius 2 and centres O and P touch each other as shown in the figure. If AD and BD are tangents, then find the length of BD

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$\boxed{\red{\bold{Given:}}}$

☆《$Radius \ of \ each \ circle = 2cm$》☆

☆《$AD \ and \ BD \ are \ are \ tangents$》☆

$\boxed{\red{\bold{Construction:}}}$

☆《Join the centre P to the point of contact and name it PC.》☆

$\boxed{\red{\bold{Solutuon:}}}$

☆ In$\underline{\triangle}$APC: ☆

$AP^2 - CP^2 = AC^2$

$(\because Pythagoras \ theorem)$

$(OA + OQ + OP)^2 - (2)^2 = AC^2$

$(2+2+2)^2 - 2^2 = AC^2$

$AC^2 = 6^2 - 2^2$

$AC^2 = 36 - 4$

$AC^2 = 32$

$AC = 16 \sqrt{2}$

☆ In$\underline{\triangle}$ ACP and$\underline{\triangle}$ ABD: ☆

$\angle ACP = \angle ABD$(each 90⁰)

$\angle CAP = \angle ABD$(common)

$By \ AA \ similarity \ criterion, \\ \triangle ACP \sim \triangle ABD$

☆ Taking side proportional: ☆

$\displaystyle{\frac{AC}{AB}} = \frac{CP}{BD}$

$\displaystyle{\frac{16 \sqrt{2}}{8}} = \frac{2}{BD}$

$BD = \displaystyle{\frac{2 × 8}{16 \sqrt{2}}}$

$BD = \displaystyle{\frac{1}{\sqrt{2}}}$

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