A and B are centres of the two circles whose radii are 5 cm and 2 cm respectively. The direct common tangents to the circles meet AB extended at P. Then P divides AB. (SSC CGL 2nd Sit. 2012)a)externally in the ratio 5 : 2b)internally in the ratio 2 : 5c)internally in the ratio 5 : 2d)externally in the ratio 7 : 2Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Here RQ is the common tangent which touches circles with centre A and B at point R and Q respectively
∴ ∠ARQ = ∠BQR = 90°
On extending the line AB, tangent RQ meet the line AB at point P.
Now, In DPBQ and DPAR,
BQ || AR, ∠P = ∠P, ∠Q = ∠R ⇒ ∠A = ∠B.
thus, DPBQ ~ DPAR {from AA theorem}

Hence, point P, divides line AB into 5 : 2 ratio externally.

Answer

Given:
- A and B are the centers of two circles with radii of 5 cm and 2 cm respectively.
- The direct common tangents to the circles meet AB extended at P.

To find:
- The ratio in which P divides AB.

Solution:
1. Let O1 and O2 be the centers of the two circles with radii 5 cm and 2 cm respectively.
2. Draw the radii OA1 and OB1 of the circles.
3. Let M be the midpoint of AB.
4. According to the properties of tangents, the line PM will be perpendicular to OA1 and OB1.
5. Therefore, triangles OAP and OBP are similar right-angled triangles.
6. Let AP = 5x and BP = 2x (assuming AP > BP).
7. In triangle OAP, using Pythagoras theorem, we have:
- (OA1)^2 = (OP)^2 + (AP)^2
- (5)^2 = (OP)^2 + (5x)^2
- (OP)^2 = 25 - 25x^2

8. In triangle OBP, using Pythagoras theorem, we have:
- (OB1)^2 = (OP)^2 + (BP)^2
- (2)^2 = (OP)^2 + (2x)^2
- (OP)^2 = 4 - 4x^2

9. Equating the values of (OP)^2 from equations (7) and (8), we get:
- 25 - 25x^2 = 4 - 4x^2
- 21x^2 = 21
- x^2 = 1
- x = 1 (since x > 0)

10. Therefore, AP = 5x = 5 and BP = 2x = 2.

Conclusion:
- P divides AB externally in the ratio 5:2.

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