Question :
In the given figure, two identical circles of radius 4 cm touch each other. A and B are the centres of the two circles. If RQ is tangent to the circle, then what is the length (in cm) of RQ?
Option 1: $3\sqrt{3}$
Option 2: $2\sqrt{6}$
Option 3: $4\sqrt{2}$
Option 4: $6\sqrt{2}$

Question

Question : Two circles touch each other externally. The radius of the first circle with centre O is 6 cm. The radius of the second circle with centre P is 3 cm. Find the length of their common tangent AB.
Option 1: $3\sqrt{2}$ cm
Option 2: $4\sqrt{2}$ cm
Option 3: $6\sqrt{3}$ cm
Option 4: $6\sqrt{2}$ cm

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Answer 1

Correct Answer: $4\sqrt{2}$

Solution :

Let RQ = RN = $x$ (because the lengths of tangents drawn from an external point to a circle are equal) and $\angle QPR=\theta$
Construction: Draw a line from A to S, such that AS $\perp$ RP (tangent at any point of a circle is perpendicular to the radius through the point of contact)
In $\triangle ANP$,
$⇒\tan \theta=\frac{AN}{PN}$
$⇒\tan \theta=\frac{4}{PN}$....................(i)
In $\triangle RQP$,
$⇒\tan \theta=\frac{RQ}{QP}$
$⇒\tan \theta=\frac{x}{QP}$
$⇒\tan \theta=\frac{x}{16}$.......................(ii)
From equation (i) and (ii)
$⇒\frac{4}{PN}=\frac{x}{16}$................(iii)
In $\triangle ANP$,
$AP^2=AN^2+PN^2$
$⇒12^2=4^2+PN^2$
$⇒PN=8\sqrt2\;\mathrm{cm}$
From equation (iii)
$⇒\frac{4}{PN}=\frac{x}{16}$
$⇒\frac{4}{8\sqrt2}=\frac{x}{16}$
$⇒x=4\sqrt2\;\mathrm{cm}$
Hence, the correct answer is $4\sqrt2$.