Two circles with centres P and Q touch each other externally at a point T. A straight line is drawn through T and meets the circle with centre P again at A and the circle with centre Q again at B. Which of the following statements about segments AP and BQ is true?

Introduction / Context:
This geometry problem deals with two circles touching externally and a line passing through their common point of contact. It tests understanding of homothety (enlargement from a point), symmetry, and how chords and radii behave when circles touch each other externally.

Given Data / Assumptions:

• Two circles have centres P and Q and touch externally at point T.
• A straight line through T meets the first circle (centre P) again at A and the second circle (centre Q) again at B.
• Segments AP and BQ join each point on the line back to the corresponding centre.
• We must identify the correct relation between AP and BQ.

Concept / Approach:
When two circles touch externally, there is a homothety (scaling transformation) centred at the point of contact T that maps one circle onto the other. This transformation sends centre P to centre Q and any point A on the first circle to a corresponding point B on the second circle along the same ray from T. Therefore, line segments TA and TB are collinear and their distances from T remain in a fixed ratio. Analysing the geometry with this symmetry shows that AP and BQ lie on the same straight line but in opposite directions, which makes them parallel (collinear). Lengths are related to the radii, but not necessarily equal unless the circles are congruent.

Step-by-Step Solution:
Step 1: Because the circles touch externally at T, line PQ passes through T and PT and QT are radii. Step 2: Consider any line through T. It cuts the circle with centre P at T and A, and the circle with centre Q at T and B. Step 3: There is a homothety with centre T that maps the first circle to the second, sending P to Q and A to B along the same line. Step 4: Under this transformation, segment AP in the first circle corresponds in direction to segment BQ in the second circle. Step 5: Vector analysis shows that AP and BQ are scalar multiples of each other with opposite sign, which means they lie on the same straight line but point in opposite directions. Step 6: Hence AP and BQ are parallel (collinear), although generally not equal in length.

Verification / Alternative Check:
A coordinate geometry approach confirms this. By placing T at the origin and P and Q on the x axis with appropriate distances, one can parametrise the line through T and find coordinates for A and B. Computing vectors AP and BQ shows that one is a negative scalar multiple of the other, which proves they are parallel for every possible line through T, not just a particular one.

Why Other Options Are Wrong:
AP = BQ would only hold if the radii of the two circles are equal, which is not stated. AP ⟂ BQ is incorrect because the vectors are collinear, not perpendicular. AP = 1⁄2 BQ is also not generally true; the scale factor depends on the ratio of the radii and is not fixed at one half.

Common Pitfalls:
Students often assume that because the circles touch, some special angle such as a right angle must appear between certain segments, but that is not the case here. Another common mistake is to focus only on one particular line through T instead of recognising that the relation must hold for every such line. Thinking in terms of homothety around the point of contact T helps reveal the true relationship between AP and BQ.

Final Answer:
The correct statement is that AP ∥ BQ (AP is parallel to BQ).