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PQ is a direct common tangent of two circles of radii r1 and r2 touching each other externally at A. Then the value of PQ2 is (SSC CGL 1st Sit. 2012)
- a)r1r2
- b)2r1r2
- c)3r1r2
- d)4r1r2
Correct answer is option 'D'. Can you explain this answer?
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PQ is a direct common tangent of two circles of radii r1 and r2 touchi...
PQ2 = (r1 + r2)2 – (r1 – r2)2 = 4r1r2
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PQ is a direct common tangent of two circles of radii r1 and r2 touchi...
Concept:
When a line is a direct common tangent to two circles, it passes through the point of contact of the two circles. This forms a right-angled triangle with the radii of the two circles as legs and the tangent line as the hypotenuse.
Explanation:
1. Let the centers of the two circles be O1 and O2, and the point of contact be A.
2. Since PQ is a direct common tangent, it passes through point A.
3. Let the distance between the centers O1 and O2 be d.
4. Therefore, the length of PQ = d - r1 - r2 (distance between centers minus sum of radii).
5. In right triangle OAQ, we have:
OA = r1, OQ = d - r1
6. Applying Pythagoras theorem in triangle OAQ:
(d - r1)^2 = r1^2 + AQ^2
d^2 - 2dr1 + r1^2 = r1^2 + AQ^2
AQ^2 = d^2 - 2dr1
7. Similarly, in right triangle OAQ2, we have:
OA2 = r2, OQ2 = d + r2
8. Applying Pythagoras theorem in triangle OAQ2:
(d + r2)^2 = r2^2 + AQ2^2
d^2 + 2dr2 + r2^2 = r2^2 + AQ2^2
AQ2^2 = d^2 + 2dr2
9. Adding equations (6) and (8):
AQ^2 + AQ2^2 = d^2 - 2dr1 + d^2 + 2dr2
PQ^2 = 2d^2 - 2dr1 + 2dr2
PQ^2 = 2(d^2 - r1 + r2)
PQ^2 = 2d^2 - 2(r1 + r2)d + 2r1r2
PQ^2 = (d - r1 - r2)^2 + 2r1r2
PQ^2 = PQ^2 + 2r1r2
PQ^2 = 4r1r2
Therefore, the value of PQ^2 is 4r1r2. Hence, option 'D' is the correct answer.
When a line is a direct common tangent to two circles, it passes through the point of contact of the two circles. This forms a right-angled triangle with the radii of the two circles as legs and the tangent line as the hypotenuse.
Explanation:
1. Let the centers of the two circles be O1 and O2, and the point of contact be A.
2. Since PQ is a direct common tangent, it passes through point A.
3. Let the distance between the centers O1 and O2 be d.
4. Therefore, the length of PQ = d - r1 - r2 (distance between centers minus sum of radii).
5. In right triangle OAQ, we have:
OA = r1, OQ = d - r1
6. Applying Pythagoras theorem in triangle OAQ:
(d - r1)^2 = r1^2 + AQ^2
d^2 - 2dr1 + r1^2 = r1^2 + AQ^2
AQ^2 = d^2 - 2dr1
7. Similarly, in right triangle OAQ2, we have:
OA2 = r2, OQ2 = d + r2
8. Applying Pythagoras theorem in triangle OAQ2:
(d + r2)^2 = r2^2 + AQ2^2
d^2 + 2dr2 + r2^2 = r2^2 + AQ2^2
AQ2^2 = d^2 + 2dr2
9. Adding equations (6) and (8):
AQ^2 + AQ2^2 = d^2 - 2dr1 + d^2 + 2dr2
PQ^2 = 2d^2 - 2dr1 + 2dr2
PQ^2 = 2(d^2 - r1 + r2)
PQ^2 = 2d^2 - 2(r1 + r2)d + 2r1r2
PQ^2 = (d - r1 - r2)^2 + 2r1r2
PQ^2 = PQ^2 + 2r1r2
PQ^2 = 4r1r2
Therefore, the value of PQ^2 is 4r1r2. Hence, option 'D' is the correct answer.
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PQ is a direct common tangent of two circles of radii r1 and r2 touching each other externally at A. Then the value of PQ2 is (SSC CGL 1st Sit. 2012)a)r1r2b)2r1r2c)3r1r2d)4r1r2Correct answer is option 'D'. Can you explain this answer?
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