A circle of radius 4 cm is externally tangent to a circle of radius 9 cm. A line is tangent to both circles and touches them at points A and B. Find the length AB? (in cm)a)12b)14c)10d)8Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Problem statement:
Find the length AB of a line that is tangent to two circles of radii 4 cm and 9 cm respectively and touches them at points A and B.

Solution:
Let O and O' be the centers of the circles of radii 4 cm and 9 cm respectively, and let C be the point of intersection of the line AB with the segment OO'.

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Since the line AB is tangent to both circles, we have

∠OAB = 90° and ∠O'BA = 90°.

Therefore, ∠AOB = ∠BO'C.

Also, since OA and OB are radii of the circle with center O, we have OA = OB = 4 cm.

Similarly, since OB and OC are radii of the circle with center O', we have OB = OC = 9 cm.

Let x be the length of AB. Then, we have

AC = AO + OC = 4 cm + 9 cm = 13 cm

and

BC = BO' - CO = 9 cm - 4 cm = 5 cm.

Applying the Pythagorean theorem to the right triangles AOC and BOC, we get

AC² = AO² + OC²
⇒ 169 = 16 + OC²
⇒ OC² = 153
⇒ OC = √153.

and

BC² = BO'² - CO²
⇒ 25 = 81 - CO²
⇒ CO² = 56
⇒ CO = √56.

Since ∠AOB = ∠BO'C, we have

∠AOC + ∠BO'C = 180°
⇒ ∠AOC = 180° - ∠BO'C
⇒ ∠AOC = 180° - ∠AOB.

Using the Law of Cosines in triangle AOC, we get

x² = AC² + AO² - 2(AC)(AO)cos∠AOC
⇒ x² = 169 + 16 - 2(13)(4)cos(180° - ∠AOB)
⇒ x² = 185 + 104cos∠AOB.

Using the Law of Cosines in triangle BOC, we get

x² = BC² + BO'² - 2(BC)(BO')cos∠BO'C
⇒ x² = 25 + 81 - 2(5)(9)cos∠AOB
⇒ x² = 106 - 90cos∠AOB.

Equating the two expressions for x², we get

185 + 104cos∠AOB = 106 - 90cos∠AOB
⇒ 194cos∠AOB = -79
⇒ cos∠AOB = -79/194.

Since ∠AOB is acute, we have

cos∠AOB = AB/OB
⇒ AB = OBcos∠AOB
⇒ AB = 9(-79/194)
⇒ AB = -36/97.

But AB is a length, so we take the absolute value to get

AB = 36/97.

Therefore, the length AB of the tangent

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