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Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ is
- a)PT = 2PQ
- b)PT < PQ
- c)PT > PQ
- d)PT = PQ
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Two circles intersect at A and B. P is a point on produced BA. PT and ...
We know that,
Property- If MNO is a secant to a circle intersecting the circle at M and N and OT is a tangent segment, then
OM × ON = OT2
From the diagram,
PAB is a secant to a circle with center O1 intersecting the circle at A and B and PT is a tangent segment,
∴ PA × PB = PT2 ……. (1)
Again, PAB is a secant to a circle with center O2 intersecting the circle at A and B and PQ is a tangent segment,
∴ PA × PB = PQ2 ……. (2)
From (1) and (2)
PT2 = PQ2
⇒ PT = PQ
Most Upvoted Answer
Two circles intersect at A and B. P is a point on produced BA. PT and ...
We can use the power of a point theorem to find the relation between PT and PQ.
Let O1 and O2 be the centers of the circles. Then, we have:
AP^2 = AB * AP = (AO1 + BO1) * (AO2 - BO2)
AQ^2 = AB * AQ = (AO1 - BO1) * (AO2 + BO2)
Note that AO1 = AO2 (since they are radii of the same circle) and BO1 = BO2 (since they are radii of the other circle).
Adding the two equations, we get:
AP^2 + AQ^2 = 2AO1 * AO2
Now, let's consider the right triangles APT and APQ. Since PT and PQ are tangents to the circles, we have:
AT = AQ and AP = AP (common side and radii of the same circle)
AP > PT and AP > PQ (since PT and PQ are tangents)
Therefore, we have:
PT^2 = AP^2 - AT^2
PQ^2 = AP^2 - AQ^2
Substituting AT = AQ into the second equation and adding the two equations, we get:
PT^2 + PQ^2 = 2AP^2 - 2AO1 * AO2
Using the first equation above, we can simplify this to:
PT^2 + PQ^2 = 2AP^2 - (AP^2 + AQ^2)
PT^2 + PQ^2 = AP^2 - AQ^2
Substituting AQ^2 = AP^2 - 2AO1 * AO2 from the first equation, we get:
PT^2 + PQ^2 = 2AP^2 - 2AO1 * AO2
Comparing this with the earlier equation for AP^2 + AQ^2, we see that:
PT^2 + PQ^2 = 2(AP^2 + AQ^2) - 4AO1 * AO2
PT^2 + PQ^2 = 2(2AO1 * AO2) - 4AO1 * AO2
PT^2 + PQ^2 = 0
Therefore, we have:
PT^2 = -PQ^2
PT = i * PQ
where i is the imaginary unit.
Thus, the relation between PT and PQ is:
PT = i * PQ
Let O1 and O2 be the centers of the circles. Then, we have:
AP^2 = AB * AP = (AO1 + BO1) * (AO2 - BO2)
AQ^2 = AB * AQ = (AO1 - BO1) * (AO2 + BO2)
Note that AO1 = AO2 (since they are radii of the same circle) and BO1 = BO2 (since they are radii of the other circle).
Adding the two equations, we get:
AP^2 + AQ^2 = 2AO1 * AO2
Now, let's consider the right triangles APT and APQ. Since PT and PQ are tangents to the circles, we have:
AT = AQ and AP = AP (common side and radii of the same circle)
AP > PT and AP > PQ (since PT and PQ are tangents)
Therefore, we have:
PT^2 = AP^2 - AT^2
PQ^2 = AP^2 - AQ^2
Substituting AT = AQ into the second equation and adding the two equations, we get:
PT^2 + PQ^2 = 2AP^2 - 2AO1 * AO2
Using the first equation above, we can simplify this to:
PT^2 + PQ^2 = 2AP^2 - (AP^2 + AQ^2)
PT^2 + PQ^2 = AP^2 - AQ^2
Substituting AQ^2 = AP^2 - 2AO1 * AO2 from the first equation, we get:
PT^2 + PQ^2 = 2AP^2 - 2AO1 * AO2
Comparing this with the earlier equation for AP^2 + AQ^2, we see that:
PT^2 + PQ^2 = 2(AP^2 + AQ^2) - 4AO1 * AO2
PT^2 + PQ^2 = 2(2AO1 * AO2) - 4AO1 * AO2
PT^2 + PQ^2 = 0
Therefore, we have:
PT^2 = -PQ^2
PT = i * PQ
where i is the imaginary unit.
Thus, the relation between PT and PQ is:
PT = i * PQ
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Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer?
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Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer?.
Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two circles intersect at A and B. P is a point on produced BA. PT and PQ are tangents to the circles. The relation of PT and PQ isa)PT = 2PQb)PT < PQc)PT > PQd)PT = PQCorrect answer is option 'D'. Can you explain this answer?.
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