Question

Two circles of radii $2and3cm$ touch each other externally. The length of direct common tangent to the two circles will be

• A. $2\sqrt{6}cm$
• B. $\sqrt{26}cm$
• C. $5cm$
• D. $2.4cm$

Answer 1

The correct option is A $2\sqrt{6}cm$

$Solution$- We drop perpendicular AR from A to BQ. AP & BQ are radii of the circles through the points of contact of PQ with the circles.
$\therefore$ AP & BQ are perpendiculars to PQ since the radius of a circle, through the point of contact of a tangent to the circle, makes ${90}^o$ angle with the same tangent.
$\therefore AP\parallel BQ......\left(i\right)$ and $\angle APQ=\angle RQP={90}^o......\left(ii\right)$.
Again $AR\perp BQ$ & $PQ\perp BQ.$
$\therefore PQ\parallel AR.......\left(iii\right)$
So, from (i), (ii) & (iii) its evident that APQR is a rectangle.
$\therefore PQ=AR$ & $AP=RQ.......\left(iv\right)$.
Now AB=sum of the radii of the two circles since the distance between the centres of two circles, who touch externally, is the sum of the radii of the two circles. So $AB=(2+3)cm=5cm$. Also $BR=BQ-RQ=(3-2)cm=1cm$ (from iv).
$\therefore$ Considering $\Delta ARB$, which is a right one,
we have $AR=\sqrt{{AB}^2-{BR}^2}=\sqrt{5^2-1^2}cm=2\sqrt{6}cm.$
$\therefore PQ=AR=2\sqrt{6}cm.$ (from iv).
Ans-Option A.