Two circles of radii 9.0 units and 4.0 units touch each other externally as in the figure. Then the length (in units) of their common tangent AB is

Correct option is C

Given:
Radius of larger circle = 9 units
Radius of smaller circle = 4 units
Circles touch externally

Formula Used:
Length of common external tangent between two circles: ​$$\sqrt{(r_1 + r_2)^2 - (r_1 - r_2)^2}$$

Solution:

Let the two circles with centers A and B and radii 4 cm and 9 cm respectively touch each other externally at a point C.
Then,
AB = AC + CB = (4 + 9) cm = 13 cm

Let PQ be a direct common tangent to the two circles.
Join AP and BQ. Then, AP ⊥ PQ and BQ ⊥ PQ.
[Radius through point of contact is perpendicular to the tangent]

Draw BL ⊥ AP. Then, PLBQ is a rectangle.
Now, LP = BQ = 4 cm and PQ = BL.
AL = (AP − LP) = (9 − 4) = 5 cm

From right triangle ALB, we have:
AB² = AL² + BL² => BL² = AB² − AL²

​$$\text{Length} = \sqrt{(9 + 4)^2 - (9 - 4)^2} \\= \sqrt{13^2 - 5^2} \\= \sqrt{169 - 25} \\= \sqrt{144} \\= 12$$

So, PQ = BL = 12 cm

Hence, the length of direct common tangent is 12 cm.​Final Answer: (c) 12